35:I[651451,["/_next/static/chunks/04p1kseqhap6_.js?dpl=nogit-1791231897","/_next/static/chunks/2_6olli-sutu3.js?dpl=nogit-1791231897","/_next/static/chunks/0zcakt1-boe_f.js?dpl=nogit-1791231897","/_next/static/chunks/0psy-n5us57a4.js?dpl=nogit-1791231897","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1791231897","/_next/static/chunks/2x_dbpl3b_l2r.js?dpl=nogit-1791231897","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1791231897","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1791231897","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1791231897","/_next/static/chunks/2othipd-ghhsj.js?dpl=nogit-1791231897","/_next/static/chunks/2gjyu-jrwu-fu.js?dpl=nogit-1791231897","/_next/static/chunks/38w1x0vrld4-w.js?dpl=nogit-1791231897","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1791231897","/_next/static/chunks/3kzt2piec6aoa.js?dpl=nogit-1791231897","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1791231897","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1791231897","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1791231897","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1791231897","/_next/static/chunks/10rmxw_yiwj40.js?dpl=nogit-1791231897","/_next/static/chunks/40lyu17d590zu.js?dpl=nogit-1791231897","/_next/static/chunks/2g-t2n9ahs3tn.js?dpl=nogit-1791231897","/_next/static/chunks/1g1suj21ijg7y.js?dpl=nogit-1791231897","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1791231897","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1791231897","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1791231897","/_next/static/chunks/2oijhb0tdir9b.js?dpl=nogit-1791231897","/_next/static/chunks/2eanemqjnv2vr.js?dpl=nogit-1791231897","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1791231897","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1791231897","/_next/static/chunks/3kxd2t9_7z5v9.js?dpl=nogit-1791231897","/_next/static/chunks/342zly86j2n91.js?dpl=nogit-1791231897","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1791231897","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1791231897","/_next/static/chunks/3hk191jki_10r.js?dpl=nogit-1791231897","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1791231897","/_next/static/chunks/2pybezgpj0ix9.js?dpl=nogit-1791231897","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1791231897","/_next/static/chunks/2eycy5cqzjdqf.js?dpl=nogit-1791231897","/_next/static/chunks/1h-hlqpew4bkg.js?dpl=nogit-1791231897","/_next/static/chunks/2ryoj58czm802.js?dpl=nogit-1791231897","/_next/static/chunks/20_c60cgo1g9g.js?dpl=nogit-1791231897","/_next/static/chunks/0w7ow4sfu0tl_.js?dpl=nogit-1791231897","/_next/static/chunks/3oiyawxg2v43g.js?dpl=nogit-1791231897","/_next/static/chunks/3q5gc8lbwr4b4.js?dpl=nogit-1791231897","/_next/static/chunks/0ffg_58gi1wsv.js?dpl=nogit-1791231897","/_next/static/chunks/24srd-xd1m5z8.js?dpl=nogit-1791231897","/_next/static/chunks/2wbhqlg6ozym8.js?dpl=nogit-1791231897","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1791231897","/_next/static/chunks/36nndwkz7iik3.js?dpl=nogit-1791231897","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1791231897","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1791231897","/_next/static/chunks/30w6c7jj-mxyi.js?dpl=nogit-1791231897","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1791231897","/_next/static/chunks/0wl6wuakgnlec.js?dpl=nogit-1791231897","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1791231897","/_next/static/chunks/2zts3qwotnhdc.js?dpl=nogit-1791231897","/_next/static/chunks/0a57o9qvnv-nb.js?dpl=nogit-1791231897","/_next/static/chunks/03u7f9lenppb3.js?dpl=nogit-1791231897","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1791231897","/_next/static/chunks/0i7rq8dl9w9qv.js?dpl=nogit-1791231897","/_next/static/chunks/0cc-_ibiuy6ux.js?dpl=nogit-1791231897","/_next/static/chunks/2nnu0qirjv_iw.js?dpl=nogit-1791231897"],"UserProgressProvider"] 36:I[196793,["/_next/static/chunks/04p1kseqhap6_.js?dpl=nogit-1791231897","/_next/static/chunks/2_6olli-sutu3.js?dpl=nogit-1791231897","/_next/static/chunks/0zcakt1-boe_f.js?dpl=nogit-1791231897","/_next/static/chunks/0psy-n5us57a4.js?dpl=nogit-1791231897","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1791231897","/_next/static/chunks/2x_dbpl3b_l2r.js?dpl=nogit-1791231897","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1791231897","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1791231897","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1791231897","/_next/static/chunks/2othipd-ghhsj.js?dpl=nogit-1791231897","/_next/static/chunks/2gjyu-jrwu-fu.js?dpl=nogit-1791231897","/_next/static/chunks/38w1x0vrld4-w.js?dpl=nogit-1791231897","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1791231897","/_next/static/chunks/3kzt2piec6aoa.js?dpl=nogit-1791231897","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1791231897","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1791231897","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1791231897","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1791231897","/_next/static/chunks/10rmxw_yiwj40.js?dpl=nogit-1791231897","/_next/static/chunks/40lyu17d590zu.js?dpl=nogit-1791231897","/_next/static/chunks/2g-t2n9ahs3tn.js?dpl=nogit-1791231897","/_next/static/chunks/1g1suj21ijg7y.js?dpl=nogit-1791231897","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1791231897","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1791231897","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1791231897","/_next/static/chunks/2oijhb0tdir9b.js?dpl=nogit-1791231897","/_next/static/chunks/2eanemqjnv2vr.js?dpl=nogit-1791231897","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1791231897","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1791231897","/_next/static/chunks/3kxd2t9_7z5v9.js?dpl=nogit-1791231897","/_next/static/chunks/342zly86j2n91.js?dpl=nogit-1791231897","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1791231897","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1791231897","/_next/static/chunks/3hk191jki_10r.js?dpl=nogit-1791231897","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1791231897","/_next/static/chunks/2pybezgpj0ix9.js?dpl=nogit-1791231897","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1791231897","/_next/static/chunks/2eycy5cqzjdqf.js?dpl=nogit-1791231897","/_next/static/chunks/1h-hlqpew4bkg.js?dpl=nogit-1791231897","/_next/static/chunks/2ryoj58czm802.js?dpl=nogit-1791231897","/_next/static/chunks/20_c60cgo1g9g.js?dpl=nogit-1791231897","/_next/static/chunks/0w7ow4sfu0tl_.js?dpl=nogit-1791231897","/_next/static/chunks/3oiyawxg2v43g.js?dpl=nogit-1791231897","/_next/static/chunks/3q5gc8lbwr4b4.js?dpl=nogit-1791231897","/_next/static/chunks/0ffg_58gi1wsv.js?dpl=nogit-1791231897","/_next/static/chunks/24srd-xd1m5z8.js?dpl=nogit-1791231897","/_next/static/chunks/2wbhqlg6ozym8.js?dpl=nogit-1791231897","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1791231897","/_next/static/chunks/36nndwkz7iik3.js?dpl=nogit-1791231897","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1791231897","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1791231897","/_next/static/chunks/30w6c7jj-mxyi.js?dpl=nogit-1791231897","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1791231897","/_next/static/chunks/0wl6wuakgnlec.js?dpl=nogit-1791231897","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1791231897","/_next/static/chunks/2zts3qwotnhdc.js?dpl=nogit-1791231897","/_next/static/chunks/0a57o9qvnv-nb.js?dpl=nogit-1791231897","/_next/static/chunks/03u7f9lenppb3.js?dpl=nogit-1791231897","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1791231897","/_next/static/chunks/0i7rq8dl9w9qv.js?dpl=nogit-1791231897","/_next/static/chunks/0cc-_ibiuy6ux.js?dpl=nogit-1791231897","/_next/static/chunks/2nnu0qirjv_iw.js?dpl=nogit-1791231897"],"Gingerbread"] 38:I[716297,["/_next/static/chunks/04p1kseqhap6_.js?dpl=nogit-1791231897","/_next/static/chunks/2_6olli-sutu3.js?dpl=nogit-1791231897","/_next/static/chunks/0zcakt1-boe_f.js?dpl=nogit-1791231897","/_next/static/chunks/0psy-n5us57a4.js?dpl=nogit-1791231897","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1791231897","/_next/static/chunks/2x_dbpl3b_l2r.js?dpl=nogit-1791231897","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1791231897","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1791231897","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1791231897","/_next/static/chunks/2othipd-ghhsj.js?dpl=nogit-1791231897","/_next/static/chunks/2gjyu-jrwu-fu.js?dpl=nogit-1791231897","/_next/static/chunks/38w1x0vrld4-w.js?dpl=nogit-1791231897","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1791231897","/_next/static/chunks/3kzt2piec6aoa.js?dpl=nogit-1791231897","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1791231897","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1791231897","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1791231897","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1791231897","/_next/static/chunks/10rmxw_yiwj40.js?dpl=nogit-1791231897","/_next/static/chunks/40lyu17d590zu.js?dpl=nogit-1791231897","/_next/static/chunks/2g-t2n9ahs3tn.js?dpl=nogit-1791231897","/_next/static/chunks/1g1suj21ijg7y.js?dpl=nogit-1791231897","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1791231897","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1791231897","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1791231897","/_next/static/chunks/2oijhb0tdir9b.js?dpl=nogit-1791231897","/_next/static/chunks/2eanemqjnv2vr.js?dpl=nogit-1791231897","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1791231897","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1791231897","/_next/static/chunks/3kxd2t9_7z5v9.js?dpl=nogit-1791231897","/_next/static/chunks/342zly86j2n91.js?dpl=nogit-1791231897","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1791231897","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1791231897","/_next/static/chunks/3hk191jki_10r.js?dpl=nogit-1791231897","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1791231897","/_next/static/chunks/2pybezgpj0ix9.js?dpl=nogit-1791231897","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1791231897","/_next/static/chunks/2eycy5cqzjdqf.js?dpl=nogit-1791231897","/_next/static/chunks/1h-hlqpew4bkg.js?dpl=nogit-1791231897","/_next/static/chunks/2ryoj58czm802.js?dpl=nogit-1791231897","/_next/static/chunks/20_c60cgo1g9g.js?dpl=nogit-1791231897","/_next/static/chunks/0w7ow4sfu0tl_.js?dpl=nogit-1791231897","/_next/static/chunks/3oiyawxg2v43g.js?dpl=nogit-1791231897","/_next/static/chunks/3q5gc8lbwr4b4.js?dpl=nogit-1791231897","/_next/static/chunks/0ffg_58gi1wsv.js?dpl=nogit-1791231897","/_next/static/chunks/24srd-xd1m5z8.js?dpl=nogit-1791231897","/_next/static/chunks/2wbhqlg6ozym8.js?dpl=nogit-1791231897","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1791231897","/_next/static/chunks/36nndwkz7iik3.js?dpl=nogit-1791231897","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1791231897","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1791231897","/_next/static/chunks/30w6c7jj-mxyi.js?dpl=nogit-1791231897","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1791231897","/_next/static/chunks/0wl6wuakgnlec.js?dpl=nogit-1791231897","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1791231897","/_next/static/chunks/2zts3qwotnhdc.js?dpl=nogit-1791231897","/_next/static/chunks/0a57o9qvnv-nb.js?dpl=nogit-1791231897","/_next/static/chunks/03u7f9lenppb3.js?dpl=nogit-1791231897","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1791231897","/_next/static/chunks/0i7rq8dl9w9qv.js?dpl=nogit-1791231897","/_next/static/chunks/0cc-_ibiuy6ux.js?dpl=nogit-1791231897","/_next/static/chunks/2nnu0qirjv_iw.js?dpl=nogit-1791231897"],"TopicRecentlyViewed"] 39:I[647935,["/_next/static/chunks/04p1kseqhap6_.js?dpl=nogit-1791231897","/_next/static/chunks/2_6olli-sutu3.js?dpl=nogit-1791231897","/_next/static/chunks/0zcakt1-boe_f.js?dpl=nogit-1791231897","/_next/static/chunks/0psy-n5us57a4.js?dpl=nogit-1791231897","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1791231897","/_next/static/chunks/2x_dbpl3b_l2r.js?dpl=nogit-1791231897","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1791231897","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1791231897","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1791231897","/_next/static/chunks/2othipd-ghhsj.js?dpl=nogit-1791231897","/_next/static/chunks/2gjyu-jrwu-fu.js?dpl=nogit-1791231897","/_next/static/chunks/38w1x0vrld4-w.js?dpl=nogit-1791231897","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1791231897","/_next/static/chunks/3kzt2piec6aoa.js?dpl=nogit-1791231897","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1791231897","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1791231897","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1791231897","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1791231897","/_next/static/chunks/10rmxw_yiwj40.js?dpl=nogit-1791231897","/_next/static/chunks/40lyu17d590zu.js?dpl=nogit-1791231897","/_next/static/chunks/2g-t2n9ahs3tn.js?dpl=nogit-1791231897","/_next/static/chunks/1g1suj21ijg7y.js?dpl=nogit-1791231897","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1791231897","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1791231897","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1791231897","/_next/static/chunks/2oijhb0tdir9b.js?dpl=nogit-1791231897","/_next/static/chunks/2eanemqjnv2vr.js?dpl=nogit-1791231897","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1791231897","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1791231897","/_next/static/chunks/3kxd2t9_7z5v9.js?dpl=nogit-1791231897","/_next/static/chunks/342zly86j2n91.js?dpl=nogit-1791231897","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1791231897","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1791231897","/_next/static/chunks/3hk191jki_10r.js?dpl=nogit-1791231897","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1791231897","/_next/static/chunks/2pybezgpj0ix9.js?dpl=nogit-1791231897","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1791231897","/_next/static/chunks/2eycy5cqzjdqf.js?dpl=nogit-1791231897","/_next/static/chunks/1h-hlqpew4bkg.js?dpl=nogit-1791231897","/_next/static/chunks/2ryoj58czm802.js?dpl=nogit-1791231897","/_next/static/chunks/20_c60cgo1g9g.js?dpl=nogit-1791231897","/_next/static/chunks/0w7ow4sfu0tl_.js?dpl=nogit-1791231897","/_next/static/chunks/3oiyawxg2v43g.js?dpl=nogit-1791231897","/_next/static/chunks/3q5gc8lbwr4b4.js?dpl=nogit-1791231897","/_next/static/chunks/0ffg_58gi1wsv.js?dpl=nogit-1791231897","/_next/static/chunks/24srd-xd1m5z8.js?dpl=nogit-1791231897","/_next/static/chunks/2wbhqlg6ozym8.js?dpl=nogit-1791231897","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1791231897","/_next/static/chunks/36nndwkz7iik3.js?dpl=nogit-1791231897","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1791231897","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1791231897","/_next/static/chunks/30w6c7jj-mxyi.js?dpl=nogit-1791231897","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1791231897","/_next/static/chunks/0wl6wuakgnlec.js?dpl=nogit-1791231897","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1791231897","/_next/static/chunks/2zts3qwotnhdc.js?dpl=nogit-1791231897","/_next/static/chunks/0a57o9qvnv-nb.js?dpl=nogit-1791231897","/_next/static/chunks/03u7f9lenppb3.js?dpl=nogit-1791231897","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1791231897","/_next/static/chunks/0i7rq8dl9w9qv.js?dpl=nogit-1791231897","/_next/static/chunks/0cc-_ibiuy6ux.js?dpl=nogit-1791231897","/_next/static/chunks/2nnu0qirjv_iw.js?dpl=nogit-1791231897"],"RecordResourceResume"] 3a:I[426249,["/_next/static/chunks/04p1kseqhap6_.js?dpl=nogit-1791231897","/_next/static/chunks/2_6olli-sutu3.js?dpl=nogit-1791231897","/_next/static/chunks/0zcakt1-boe_f.js?dpl=nogit-1791231897","/_next/static/chunks/0psy-n5us57a4.js?dpl=nogit-1791231897","/_next/static/chunks/3mz725-y91ph6.js?dpl=nogit-1791231897","/_next/static/chunks/2x_dbpl3b_l2r.js?dpl=nogit-1791231897","/_next/static/chunks/13_9hjun60hj-.js?dpl=nogit-1791231897","/_next/static/chunks/2bxmml65nu49g.js?dpl=nogit-1791231897","/_next/static/chunks/0zkcxwtbsqliz.js?dpl=nogit-1791231897","/_next/static/chunks/2othipd-ghhsj.js?dpl=nogit-1791231897","/_next/static/chunks/2gjyu-jrwu-fu.js?dpl=nogit-1791231897","/_next/static/chunks/38w1x0vrld4-w.js?dpl=nogit-1791231897","/_next/static/chunks/3jpl-42mbad5n.js?dpl=nogit-1791231897","/_next/static/chunks/3kzt2piec6aoa.js?dpl=nogit-1791231897","/_next/static/chunks/19ijf68pshdcn.js?dpl=nogit-1791231897","/_next/static/chunks/2os7nxouq91eq.js?dpl=nogit-1791231897","/_next/static/chunks/3ml3vooxhiupb.js?dpl=nogit-1791231897","/_next/static/chunks/04zr-wqafvwoa.js?dpl=nogit-1791231897","/_next/static/chunks/10rmxw_yiwj40.js?dpl=nogit-1791231897","/_next/static/chunks/40lyu17d590zu.js?dpl=nogit-1791231897","/_next/static/chunks/2g-t2n9ahs3tn.js?dpl=nogit-1791231897","/_next/static/chunks/1g1suj21ijg7y.js?dpl=nogit-1791231897","/_next/static/chunks/3lz9pr5v38m4h.js?dpl=nogit-1791231897","/_next/static/chunks/2mv9l1xl_aa4j.js?dpl=nogit-1791231897","/_next/static/chunks/0v4byuex1x38h.js?dpl=nogit-1791231897","/_next/static/chunks/2oijhb0tdir9b.js?dpl=nogit-1791231897","/_next/static/chunks/2eanemqjnv2vr.js?dpl=nogit-1791231897","/_next/static/chunks/2dx-vzwyw9a3g.js?dpl=nogit-1791231897","/_next/static/chunks/0llqhg6-bxn_b.js?dpl=nogit-1791231897","/_next/static/chunks/3kxd2t9_7z5v9.js?dpl=nogit-1791231897","/_next/static/chunks/342zly86j2n91.js?dpl=nogit-1791231897","/_next/static/chunks/1ku-vdy4ggo41.js?dpl=nogit-1791231897","/_next/static/chunks/1coblm1-go81_.js?dpl=nogit-1791231897","/_next/static/chunks/3hk191jki_10r.js?dpl=nogit-1791231897","/_next/static/chunks/0k-y_4mcfub-p.js?dpl=nogit-1791231897","/_next/static/chunks/2pybezgpj0ix9.js?dpl=nogit-1791231897","/_next/static/chunks/075i0tnxt4not.js?dpl=nogit-1791231897","/_next/static/chunks/2eycy5cqzjdqf.js?dpl=nogit-1791231897","/_next/static/chunks/1h-hlqpew4bkg.js?dpl=nogit-1791231897","/_next/static/chunks/2ryoj58czm802.js?dpl=nogit-1791231897","/_next/static/chunks/20_c60cgo1g9g.js?dpl=nogit-1791231897","/_next/static/chunks/0w7ow4sfu0tl_.js?dpl=nogit-1791231897","/_next/static/chunks/3oiyawxg2v43g.js?dpl=nogit-1791231897","/_next/static/chunks/3q5gc8lbwr4b4.js?dpl=nogit-1791231897","/_next/static/chunks/0ffg_58gi1wsv.js?dpl=nogit-1791231897","/_next/static/chunks/24srd-xd1m5z8.js?dpl=nogit-1791231897","/_next/static/chunks/2wbhqlg6ozym8.js?dpl=nogit-1791231897","/_next/static/chunks/13gj59kzq50t_.js?dpl=nogit-1791231897","/_next/static/chunks/36nndwkz7iik3.js?dpl=nogit-1791231897","/_next/static/chunks/1lsxkfxy262n-.js?dpl=nogit-1791231897","/_next/static/chunks/0zg1om_k0m5zv.js?dpl=nogit-1791231897","/_next/static/chunks/30w6c7jj-mxyi.js?dpl=nogit-1791231897","/_next/static/chunks/1ksyn9m68446d.js?dpl=nogit-1791231897","/_next/static/chunks/0wl6wuakgnlec.js?dpl=nogit-1791231897","/_next/static/chunks/1hnuhh5t7cfaj.js?dpl=nogit-1791231897","/_next/static/chunks/2zts3qwotnhdc.js?dpl=nogit-1791231897","/_next/static/chunks/0a57o9qvnv-nb.js?dpl=nogit-1791231897","/_next/static/chunks/03u7f9lenppb3.js?dpl=nogit-1791231897","/_next/static/chunks/3xskuweujaw05.js?dpl=nogit-1791231897","/_next/static/chunks/0i7rq8dl9w9qv.js?dpl=nogit-1791231897","/_next/static/chunks/0cc-_ibiuy6ux.js?dpl=nogit-1791231897","/_next/static/chunks/2nnu0qirjv_iw.js?dpl=nogit-1791231897"],"TopicQuestionBank"] 26:["$","$L35",null,{"bundle":{"topicIds":["5ef91db9-d11d-45a6-b4f4-105181ee3b51"],"paperIds":[],"stageIds":[],"stageTopicIds":[],"stageAssessmentIds":[],"topicNoteIds":[]},"children":[["$","$L36",null,{"feature":"question-bank"}],"$L37",["$","$L38",null,{"subjectCode":"a_level_maths","subjectName":"Maths","topicSlug":"a-level-pure-mathematics-3-2-arithmetic-series","topicTitle":"3.2 Arithmetic Series","sectionTitle":null,"activity":"worksheet","resourceTitle":"3.2 Arithmetic Series Questions","resourceHref":"/a-level/maths/edexcel/a-level-pure-mathematics-3-2-arithmetic-series/questions/q1","saveTrail":false}],["$","$L39",null,{"href":"/a-level/maths/edexcel/a-level-pure-mathematics-3-2-arithmetic-series/questions/q1","label":"3.2 Arithmetic Series","subjectCode":"a_level_maths","surface":"questions"}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"name\":\"Question 28 – 3.2 Arithmetic Series - A Level Maths Questions\",\"description\":\"Practise 3.2 Arithmetic Series A Level Maths exam-style questions with worked solutions on MathsGenie.\",\"learningResourceType\":\"Practice questions\",\"educationalLevel\":\"A Level\",\"about\":{\"@type\":\"Thing\",\"name\":\"3.2 Arithmetic Series\"},\"url\":\"https://mathsgenie.co.uk/a-level/maths/edexcel/a-level-pure-mathematics-3-2-arithmetic-series/questions/q28\",\"inLanguage\":\"en-GB\",\"provider\":{\"@id\":\"https://mathsgenie.co.uk/#organization\"}}"}}],["$","div",null,{"className":"min-h-0 w-full flex-1 overflow-hidden bg-background","children":["$","$L3a",null,{"backHref":"/a-level/maths/edexcel/questions","basePath":"/a-level/maths/edexcel/a-level-pure-mathematics-3-2-arithmetic-series/questions","formulaUrl":"https://assets.mathsgenie.co.uk/board-resources/a-level-maths/edexcel/formula-book.pdf","initialIndex":27,"initialQuestionWindow":[{"id":"7b29e5ab-30ed-44ca-8b8f-80a7f8efa445","revisionId":"66b21db5-9f81-405f-bbdb-e21c8fd38fab","questionNumber":null,"questionType":"LA","specification":"A carbon sequestration plant is gradually scaling up its operations. On the first day of its new cycle, the plant captures 18 tonnes of carbon dioxide. Each subsequent day, the amount of carbon dioxide captured increases by 2.5 tonnes.","difficultyBand":"medium","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"8b74c9bf-cf58-4172-a772-24497711c1f8","label":"a","content":"Determine an expression for the amount of carbon dioxide captured on the $n$th day of the cycle.","markscheme":"- Identifies $a = 18$ and $d = 2.5$ and uses the formula $a + (n-1)d$ M1\n- Obtains $2.5n + 15.5$ or equivalent A1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"arithmetic_formula","label":"Using Arithmetic Sequence Formula","steps":[{"type":"text","order":0,"title":"Identify the key components","content":"To write an expression for the amount of carbon dioxide captured on the $n$th day, we can model this situation as an arithmetic sequence.\n\nFrom the question, we are given:\n- The amount captured on the first day ($a$ or $u_1$) is $18$ tonnes.\n- The daily increase ($d$, the common difference) is $2.5$ tonnes.","highlights":[{"text":"On the first day of its new cycle, the plant captures 18 tonnes","location":"specification"},{"text":"Each subsequent day, the amount of carbon dioxide captured increases by 2.5 tonnes.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"The standard formula for the $n$th term of an arithmetic sequence is:"},{"latex":"$$$u_n = \\htmlData{anchor=first-term}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=difference}{\\color{@amber}{d}}$$"},{"text":"Where:"},{"latex":"$$$\\begin{aligned} \\htmlData{anchor=first-val}{\\color{@sky}{a}} &= 18 \\\\ \\htmlData{anchor=diff-val}{\\color{@amber}{d}} &= 2.5 \\end{aligned}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"first-term"},{"color":"amber","style":"text-color","anchor":"difference"}],"connections":[{"to":"first-term","from":"first-val","color":"sky","label":"substitute a","style":"solid"},{"to":"difference","from":"diff-val","color":"amber","label":"substitute d","style":"solid"}]},"order":1,"title":"Recall the sequence formula","description":"We identify the first term and common difference to plug into the general arithmetic sequence formula."},{"type":"text","order":2,"title":"Substitute and expand","content":"Substitute $a = 18$ and $d = 2.5$ into the formula:\n\n$$u_n = 18 + (n - 1)(2.5)$$\n\nNow, expand the bracket:\n\n$$u_n = 18 + 2.5n - 2.5$$"},{"type":"text","order":3,"title":"Simplify the expression","content":"Combine the constant terms ($18 - 2.5$) to find the final simplified expression:\n\n$$u_n = 2.5n + 15.5$$\n\nThis expression gives the amount of carbon dioxide (in tonnes) captured on the $n$th day."}]},{"id":"linear_model","label":"Linear Modeling","steps":[{"type":"text","order":0,"title":"Define the model","content":"We want to model the amount of carbon dioxide captured on the $n$th day, $u_n$, as a linear function of the form:\n\n$$u_n = dn + c$$\n\nwhere:\n- $d$ is the constant daily rate of increase (the gradient),\n- $n$ is the day number,\n- $c$ is the $y$-intercept parameter to be found."},{"type":"text","order":1,"title":"Identify rate of change","content":"From the problem, the daily amount captured increases by a constant amount of $2.5$ tonnes each day. This is our rate of change, $d$:\n\n$$d = 2.5$$\n\nSubstituting this into our equation gives:\n\n$$u_n = 2.5n + c$$","highlights":[{"text":"increases by 2.5 tonnes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{On the first day (}\\text{\\htmlData{anchor=n-val}{\\color{@sky}{n = 1}}}\\text{), the capture is }\\text{\\htmlData{anchor=u-val}{\\color{@amber}{18}}}\\text{ tonnes.}$"},{"latex":"$$\\text{\\htmlData{anchor=u-sub}{\\color{@amber}{18}}} = 2.5(\\text{\\htmlData{anchor=n-sub}{\\color{@sky}{1}}}) + c$"},{"latex":"$$18 = 2.5 + c \\to c = 15.5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"amber","style":"text-color","anchor":"u-val"}],"connections":[{"to":"n-sub","from":"n-val","color":"sky","label":null,"style":"solid"},{"to":"u-sub","from":"u-val","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Solve for parameter c","description":"We substitute $n = 1$ and $u_1 = 18$ into the linear equation to solve for $c$."},{"type":"text","order":3,"title":"Write final expression","content":"Now substitute $c = 15.5$ back into our linear function to write the complete expression for $u_n$:\n\n$$u_n = 2.5n + 15.5$$"}]}],"generatedAt":"2026-06-09T05:56:40.945Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"9d041ac7-fe10-4d7c-a3d0-08ccfa4532f3","label":"bi","content":"The plant has a daily operational target of at least 150.5 tonnes. Calculate the number of days required to reach this specific target for the first time.","markscheme":"- Sets their expression from part (a) equal to $150.5$: $2.5n + 15.5 = 150.5$ M1\n- Subtracts $15.5$ and divides by $2.5$ to find $n$ M1\n- Obtains $n = 54$ days A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Formula Approach","steps":[{"type":"text","order":0,"title":"Identify sequence parameters","content":"First, let's identify the key components of our arithmetic sequence from the context:\n\n- The first term is $u_1 = 18$ (representing the amount of carbon dioxide captured on the first day).\n- The common difference is $d = 2.5$ (representing the daily increase of 2.5 tonnes).\n\nWe want to find the day $n$ when the daily operational target reaches at least $150.5$ tonnes.","highlights":[{"text":"18 tonnes","location":"specification"},{"text":"increases by 2.5 tonnes","location":"specification"},{"text":"daily operational target of at least 150.5 tonnes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_n = \\text{first term} + (n-1) \\times \\text{common difference}$"},{"latex":"$$u_n = \\htmlData{anchor=first-term}{\\color{@sky}{18}} + (n-1)\\htmlData{anchor=diff}{\\color{@amber}{2.5}}$"},{"latex":"$$150.5 = \\htmlData{anchor=first-term-sub}{\\color{@sky}{18}} + \\htmlData{anchor=diff-sub}{\\color{@amber}{2.5}}n - 2.5$"},{"latex":"$$150.5 = 2.5n + 15.5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"first-term"},{"color":"amber","style":"text-color","anchor":"diff"},{"color":"sky","style":"text-color","anchor":"first-term-sub"},{"color":"amber","style":"text-color","anchor":"diff-sub"}],"connections":[{"to":"first-term-sub","from":"first-term","color":"sky","label":"substitute u_1","style":"solid"},{"to":"diff-sub","from":"diff","color":"amber","label":"substitute d","style":"solid"}]},"order":1,"title":"Set up the equation","description":"We use the general term formula for an arithmetic sequence: $u_n = u_1 + (n-1)d$. By substituting $u_1 = 18$ and $d = 2.5$, we can find the general expression for the $n$-th day and set it equal to $150.5$."},{"type":"text","order":2,"title":"Solve for n","content":"Now, we solve for $n$ step-by-step:\n\n$$2.5n + 15.5 = 150.5$$\n\nSubtract $15.5$ from both sides to isolate the variable term:\n\n$$2.5n = 135$$\n\nDivide by $2.5$ to find $n$:\n\n$$n = \\frac{135}{2.5}$$\n\n$$n = 54$$\n\nThis calculation shows that the daily operational target is reached on exactly the $54^{\\text{th}}$ day."},{"type":"text","order":3,"title":"State final answer","content":"The number of days required to reach the target is **54** days.\n\n### Markscheme Breakdown:\n- **M1**: Correctly setting up the equation: $2.5n + 15.5 = 150.5$ (or $18 + 2.5(n-1) = 150.5$)\n- **M1**: Rearranging to isolate $n$\n- **A1**: Finding the correct value: $n = 54$"}]},{"id":"calculator","label":"Using Graphic Display Calculator","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"From the problem description, on the first day, the carbon captured is $a = 18$ tonnes, and it increases by $d = 2.5$ tonnes each day.\n\nThe general formula for the $n$-th term of an arithmetic sequence is:\n\n$$u_n = a + (n - 1)d$$\n\nSubstituting $a = 18$ and $d = 2.5$:\n\n$$u_n = 18 + 2.5(n - 1) = 2.5n + 15.5$$\n\nTo find when the daily capture reaches at least $150.5$ tonnes, we set our expression equal to $150.5$:\n\n$$2.5n + 15.5 = 150.5$$","highlights":[{"text":"daily operational target of at least 150.5 tonnes","location":"specification"}]},{"type":"text","order":1,"title":"Solve using GDC","content":"Rather than rearranging the equation manually, we can input this linear model into a Graphic Display Calculator (GDC) to find the value of $n$ where the output is exactly $150.5$. We can do this either by finding the intersection of two graphs or by using a numerical solver.","calculator":[{"gdc":{"expression":"y1 = 2.5x + 15.5, y2 = 150.5","instructions":"1. Go to the Graph menu.\n2. Enter $y1 = 2.5x + 15.5$ and $y2 = 150.5$.\n3. Adjust the view window to display both lines (e.g., $x$ up to 60, $y$ up to 160).\n4. Select 'G-Solv' (F5) then 'INTSECT' (F5) to find the intersection point."},"ti84":{"expression":"Y1 = 2.5X + 15.5, Y2 = 150.5","instructions":"1. Press [Y=] and enter $Y_1 = 2.5X + 15.5$ and $Y_2 = 150.5$.\n2. Press [WINDOW] and set $X_{\\text{min}} = 0$, $X_{\\text{max}} = 60$, $Y_{\\text{min}} = 0$, and $Y_{\\text{max}} = 160$.\n3. Press [GRAPH].\n4. Press [2nd] [CALC] and select '5: intersect'.\n5. Press [ENTER] three times to find the intersection point."},"desmos":{"expression":"y = 2.5x + 15.5","instructions":"1. In the first input bar, type $y = 2.5x + 15.5$.\n2. In the second input bar, type $y = 150.5$.\n3. Adjust the graph viewport and click directly on the intersection point of the two lines to display its coordinates."},"description":"Find the intersection point of the sequence equation and the target line."}]},{"type":"text","order":2,"title":"State the final answer","content":"Using any of these calculator methods, we find the intersection point coordinates are:\n\n$$(54, 150.5)$$\n\nThis means that when $n = 54$, the daily capture is exactly $150.5$ tonnes. \n\nTherefore, it will take **54 days** to reach the target for the first time."}]}],"generatedAt":"2026-06-09T05:59:05.170Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"b03c3998-32e7-4f0f-998a-93338f213e69","label":"bii","content":"An environmental analyst predicts that by the time the plant hits this target, the total amount of carbon dioxide sequestered during the cycle will exceed 4,500 tonnes. Determine, with a calculation, whether the analyst's prediction is correct.","markscheme":"- Uses an arithmetic series sum formula with $n = 54$, e.g., $S_{54} = \\frac{54}{2}(18 + 150.5)$ or $S_{54} = \\frac{54}{2}(2(18) + 53(2.5))$ M1\n- Evaluates to find the total sum $S_{54} = 4549.5$ A1\n- States that $4549.5 > 4500$ and concludes that the analyst is correct C1","marks":3,"order":2,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"sum_using_first_and_last_term","label":"Using First and Last Term Formula","steps":[{"type":"text","order":0,"title":"Identify the parameters","content":"To find the total amount of carbon dioxide sequestered up to the target day, we need to sum the terms of our arithmetic series.\n\nFrom the previous parts of the question, we know:\n- The number of days to reach the target is $n = 54$\n- The amount sequestered on the first day is $a = 18$ tonnes\n- The amount sequestered on the target day ($54^{\\text{th}}$ day) is the last term $l = 150.5$ tonnes","highlights":[{"text":"total amount of carbon dioxide sequestered during the cycle will exceed 4,500 tonnes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_n = \\frac{\\htmlData{anchor=n-var}{n}}{2}(\\htmlData{anchor=a-var}{a} + \\htmlData{anchor=l-var}{l})$"},{"latex":"$$S_{54} = \\frac{\\htmlData{anchor=n-val}{\\color{@sky}{54}}}{2}(\\htmlData{anchor=a-val}{\\color{@amber}{18}} + \\htmlData{anchor=l-val}{\\color{@purple}{150.5}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"amber","style":"text-color","anchor":"a-val"},{"color":"purple","style":"text-color","anchor":"l-val"}],"connections":[{"to":"n-val","from":"n-var","color":"sky","label":null,"style":"solid"},{"to":"a-val","from":"a-var","color":"amber","label":null,"style":"solid"},{"to":"l-val","from":"l-var","color":"purple","label":null,"style":"solid"}]},"order":1,"title":"Apply sum formula","description":"We use the arithmetic series sum formula $S_n = \\frac{n}{2}(a+l)$, where $a$ is the first term and $l$ is the last term."},{"type":"text","order":2,"title":"Calculate the total","content":"Now, we evaluate the sum:\n\n$$\\begin{aligned} S_{54} &= 27 \\times (18 + 150.5) \\\\ S_{54} &= 27 \\times 168.5 \\\\ S_{54} &= 4549.5 \\text{ tonnes} \\end{aligned}$$\n\nThis calculation earns the first two marks (**M1** for setting up the sum and **A1** for the correct value).","calculator":[{"gdc":{"expression":"54/2*(18+150.5)","instructions":"Type `54 / 2 * (18 + 150.5)` on the main calculation screen and press ENTER."},"ti84":{"expression":"54/2*(18+150.5)","instructions":"Type `54 / 2 * (18 + 150.5)` on the home screen and press ENTER."},"desmos":{"expression":"\\frac{54}{2}(18 + 150.5)","instructions":"Type `54/2 * (18 + 150.5)` into the expression box to see the result."},"description":"Calculate the total sum directly using basic arithmetic."}]},{"type":"text","order":3,"title":"Make the conclusion","content":"Finally, we compare our calculated sum to the threshold given in the analyst's prediction:\n\n$$4549.5 \\text{ tonnes} > 4500 \\text{ tonnes}$$\n\nSince $4549.5 > 4500$, the analyst's prediction is **correct** (**C1**).","highlights":[{"text":"Determine, with a calculation, whether the analyst's prediction is correct.","location":"specification"}]}]},{"id":"sum_using_first_term_and_difference","label":"Using First Term and Common Difference Formula","steps":[{"type":"text","order":0,"title":"Identify key parameters","content":"From the problem description, the daily carbon dioxide capture forms an arithmetic sequence with:\n* First term $a = 18$ tonnes (the amount captured on the first day)\n* Common difference $d = 2.5$ tonnes (the daily increase)\n* Number of days $n = 54$\n\nWe need to find the total amount of carbon dioxide sequestered over these $54$ days and see if it exceeds $4,500$ tonnes.","highlights":[{"text":"captures 18 tonnes of carbon dioxide","location":"specification"},{"text":"increases by 2.5 tonnes","location":"specification"},{"text":"will exceed 4,500 tonnes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_n = \\frac{\\htmlData{anchor=n-val}{\\color{@sky}{n}}}{2}\\left(2\\htmlData{anchor=a-val}{\\color{@amber}{a}} + (\\htmlData{anchor=n-sub}{\\color{@sky}{n}} - 1)\\htmlData{anchor=d-val}{\\color{@purple}{d}}\\right)$"},{"latex":"$$S_{54} = \\frac{\\color{@sky}{54}}{2}\\left(2(\\color{@amber}{18}) + (\\color{@sky}{54} - 1)(\\color{@purple}{2.5})\\right)$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"amber","style":"text-color","anchor":"a-val"},{"color":"purple","style":"text-color","anchor":"d-val"}],"connections":[{"to":"n-sub","from":"n-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Apply arithmetic sum formula","description":"We will use the sum formula of an arithmetic series $S_n = \\frac{n}{2}(2a + (n-1)d)$ to compute the total captured dioxide by substituting our values: $a = 18$, $d = 2.5$, and $n = 54$."},{"type":"text","order":2,"title":"Calculate total sum","content":"Let's calculate the value step-by-step:\n\n$$\\begin{aligned} S_{54} &= 27 \\times (36 + 53 \\times 2.5) \\\\ &= 27 \\times (36 + 132.5) \\\\ &= 27 \\times 168.5 \\\\ &= 4549.5 \\text{ tonnes} \\end{aligned}$$\n\nThis calculation earns the first **M1** mark for setting up the sum equation and the **A1** mark for correctly evaluating the sum to $4549.5$.","calculator":[{"gdc":{"expression":"54/2 * (2 * 18 + 53 * 2.5)","instructions":"Enter `54/2 * (2 * 18 + 53 * 2.5)` into the main calculator screen."},"ti84":{"expression":"54/2 * (2 * 18 + 53 * 2.5)","instructions":"Type `54/2 * (2 * 18 + 53 * 2.5)` and press ENTER."},"desmos":{"expression":"54/2 * (2 * 18 + (54 - 1) * 2.5)","instructions":"Type the equation in a single cell: `54/2 * (2 * 18 + (54 - 1) * 2.5)`."},"description":"To calculate the sum on a graphing calculator, simply evaluate the expression directly."}]},{"type":"text","order":3,"title":"Compare and conclude","content":"We compare our calculated value with the analyst's prediction:\n\n$$4549.5 \\text{ tonnes} > 4500 \\text{ tonnes}$$\n\nSince the total amount of carbon dioxide sequestered ($4549.5$ tonnes) is greater than $4500$ tonnes, we conclude that **the analyst's prediction is correct**.\n\nThis final conclusion earns the **C1** mark.","highlights":[{"text":"exceed 4,500 tonnes","location":"specification"}]}]}],"generatedAt":"2026-06-09T05:59:41.038Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"8b74c9bf-cf58-4172-a772-24497711c1f8:arithmetic_formula","label":"(a) Using Arithmetic Sequence Formula","steps":[{"type":"text","order":0,"title":"Identify the key components","content":"To write an expression for the amount of carbon dioxide captured on the $n$th day, we can model this situation as an arithmetic sequence.\n\nFrom the question, we are given:\n- The amount captured on the first day ($a$ or $u_1$) is $18$ tonnes.\n- The daily increase ($d$, the common difference) is $2.5$ tonnes.","highlights":[{"text":"On the first day of its new cycle, the plant captures 18 tonnes","location":"specification"},{"text":"Each subsequent day, the amount of carbon dioxide captured increases by 2.5 tonnes.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"The standard formula for the $n$th term of an arithmetic sequence is:"},{"latex":"$$$u_n = \\htmlData{anchor=first-term}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=difference}{\\color{@amber}{d}}$$"},{"text":"Where:"},{"latex":"$$$\\begin{aligned} \\htmlData{anchor=first-val}{\\color{@sky}{a}} &= 18 \\\\ \\htmlData{anchor=diff-val}{\\color{@amber}{d}} &= 2.5 \\end{aligned}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"first-term"},{"color":"amber","style":"text-color","anchor":"difference"}],"connections":[{"to":"first-term","from":"first-val","color":"sky","label":"substitute a","style":"solid"},{"to":"difference","from":"diff-val","color":"amber","label":"substitute d","style":"solid"}]},"order":1,"title":"Recall the sequence formula","description":"We identify the first term and common difference to plug into the general arithmetic sequence formula."},{"type":"text","order":2,"title":"Substitute and expand","content":"Substitute $a = 18$ and $d = 2.5$ into the formula:\n\n$$u_n = 18 + (n - 1)(2.5)$$\n\nNow, expand the bracket:\n\n$$u_n = 18 + 2.5n - 2.5$$"},{"type":"text","order":3,"title":"Simplify the expression","content":"Combine the constant terms ($18 - 2.5$) to find the final simplified expression:\n\n$$u_n = 2.5n + 15.5$$\n\nThis expression gives the amount of carbon dioxide (in tonnes) captured on the $n$th day."}]},{"id":"8b74c9bf-cf58-4172-a772-24497711c1f8:linear_model","label":"(a) Linear Modeling","steps":[{"type":"text","order":0,"title":"Define the model","content":"We want to model the amount of carbon dioxide captured on the $n$th day, $u_n$, as a linear function of the form:\n\n$$u_n = dn + c$$\n\nwhere:\n- $d$ is the constant daily rate of increase (the gradient),\n- $n$ is the day number,\n- $c$ is the $y$-intercept parameter to be found."},{"type":"text","order":1,"title":"Identify rate of change","content":"From the problem, the daily amount captured increases by a constant amount of $2.5$ tonnes each day. This is our rate of change, $d$:\n\n$$d = 2.5$$\n\nSubstituting this into our equation gives:\n\n$$u_n = 2.5n + c$$","highlights":[{"text":"increases by 2.5 tonnes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{On the first day (}\\text{\\htmlData{anchor=n-val}{\\color{@sky}{n = 1}}}\\text{), the capture is }\\text{\\htmlData{anchor=u-val}{\\color{@amber}{18}}}\\text{ tonnes.}$"},{"latex":"$$\\text{\\htmlData{anchor=u-sub}{\\color{@amber}{18}}} = 2.5(\\text{\\htmlData{anchor=n-sub}{\\color{@sky}{1}}}) + c$"},{"latex":"$$18 = 2.5 + c \\to c = 15.5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"amber","style":"text-color","anchor":"u-val"}],"connections":[{"to":"n-sub","from":"n-val","color":"sky","label":null,"style":"solid"},{"to":"u-sub","from":"u-val","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Solve for parameter c","description":"We substitute $n = 1$ and $u_1 = 18$ into the linear equation to solve for $c$."},{"type":"text","order":3,"title":"Write final expression","content":"Now substitute $c = 15.5$ back into our linear function to write the complete expression for $u_n$:\n\n$$u_n = 2.5n + 15.5$$"}]},{"id":"9d041ac7-fe10-4d7c-a3d0-08ccfa4532f3:algebraic","label":"(bi) Algebraic Formula Approach","steps":[{"type":"text","order":0,"title":"Identify sequence parameters","content":"First, let's identify the key components of our arithmetic sequence from the context:\n\n- The first term is $u_1 = 18$ (representing the amount of carbon dioxide captured on the first day).\n- The common difference is $d = 2.5$ (representing the daily increase of 2.5 tonnes).\n\nWe want to find the day $n$ when the daily operational target reaches at least $150.5$ tonnes.","highlights":[{"text":"18 tonnes","location":"specification"},{"text":"increases by 2.5 tonnes","location":"specification"},{"text":"daily operational target of at least 150.5 tonnes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_n = \\text{first term} + (n-1) \\times \\text{common difference}$"},{"latex":"$$u_n = \\htmlData{anchor=first-term}{\\color{@sky}{18}} + (n-1)\\htmlData{anchor=diff}{\\color{@amber}{2.5}}$"},{"latex":"$$150.5 = \\htmlData{anchor=first-term-sub}{\\color{@sky}{18}} + \\htmlData{anchor=diff-sub}{\\color{@amber}{2.5}}n - 2.5$"},{"latex":"$$150.5 = 2.5n + 15.5$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"first-term"},{"color":"amber","style":"text-color","anchor":"diff"},{"color":"sky","style":"text-color","anchor":"first-term-sub"},{"color":"amber","style":"text-color","anchor":"diff-sub"}],"connections":[{"to":"first-term-sub","from":"first-term","color":"sky","label":"substitute u_1","style":"solid"},{"to":"diff-sub","from":"diff","color":"amber","label":"substitute d","style":"solid"}]},"order":1,"title":"Set up the equation","description":"We use the general term formula for an arithmetic sequence: $u_n = u_1 + (n-1)d$. By substituting $u_1 = 18$ and $d = 2.5$, we can find the general expression for the $n$-th day and set it equal to $150.5$."},{"type":"text","order":2,"title":"Solve for n","content":"Now, we solve for $n$ step-by-step:\n\n$$2.5n + 15.5 = 150.5$$\n\nSubtract $15.5$ from both sides to isolate the variable term:\n\n$$2.5n = 135$$\n\nDivide by $2.5$ to find $n$:\n\n$$n = \\frac{135}{2.5}$$\n\n$$n = 54$$\n\nThis calculation shows that the daily operational target is reached on exactly the $54^{\\text{th}}$ day."},{"type":"text","order":3,"title":"State final answer","content":"The number of days required to reach the target is **54** days.\n\n### Markscheme Breakdown:\n- **M1**: Correctly setting up the equation: $2.5n + 15.5 = 150.5$ (or $18 + 2.5(n-1) = 150.5$)\n- **M1**: Rearranging to isolate $n$\n- **A1**: Finding the correct value: $n = 54$"}]},{"id":"9d041ac7-fe10-4d7c-a3d0-08ccfa4532f3:calculator","label":"(bi) Using Graphic Display Calculator","steps":[{"type":"text","order":0,"title":"Set up the equation","content":"From the problem description, on the first day, the carbon captured is $a = 18$ tonnes, and it increases by $d = 2.5$ tonnes each day.\n\nThe general formula for the $n$-th term of an arithmetic sequence is:\n\n$$u_n = a + (n - 1)d$$\n\nSubstituting $a = 18$ and $d = 2.5$:\n\n$$u_n = 18 + 2.5(n - 1) = 2.5n + 15.5$$\n\nTo find when the daily capture reaches at least $150.5$ tonnes, we set our expression equal to $150.5$:\n\n$$2.5n + 15.5 = 150.5$$","highlights":[{"text":"daily operational target of at least 150.5 tonnes","location":"specification"}]},{"type":"text","order":1,"title":"Solve using GDC","content":"Rather than rearranging the equation manually, we can input this linear model into a Graphic Display Calculator (GDC) to find the value of $n$ where the output is exactly $150.5$. We can do this either by finding the intersection of two graphs or by using a numerical solver.","calculator":[{"gdc":{"expression":"y1 = 2.5x + 15.5, y2 = 150.5","instructions":"1. Go to the Graph menu.\n2. Enter $y1 = 2.5x + 15.5$ and $y2 = 150.5$.\n3. Adjust the view window to display both lines (e.g., $x$ up to 60, $y$ up to 160).\n4. Select 'G-Solv' (F5) then 'INTSECT' (F5) to find the intersection point."},"ti84":{"expression":"Y1 = 2.5X + 15.5, Y2 = 150.5","instructions":"1. Press [Y=] and enter $Y_1 = 2.5X + 15.5$ and $Y_2 = 150.5$.\n2. Press [WINDOW] and set $X_{\\text{min}} = 0$, $X_{\\text{max}} = 60$, $Y_{\\text{min}} = 0$, and $Y_{\\text{max}} = 160$.\n3. Press [GRAPH].\n4. Press [2nd] [CALC] and select '5: intersect'.\n5. Press [ENTER] three times to find the intersection point."},"desmos":{"expression":"y = 2.5x + 15.5","instructions":"1. In the first input bar, type $y = 2.5x + 15.5$.\n2. In the second input bar, type $y = 150.5$.\n3. Adjust the graph viewport and click directly on the intersection point of the two lines to display its coordinates."},"description":"Find the intersection point of the sequence equation and the target line."}]},{"type":"text","order":2,"title":"State the final answer","content":"Using any of these calculator methods, we find the intersection point coordinates are:\n\n$$(54, 150.5)$$\n\nThis means that when $n = 54$, the daily capture is exactly $150.5$ tonnes. \n\nTherefore, it will take **54 days** to reach the target for the first time."}]},{"id":"b03c3998-32e7-4f0f-998a-93338f213e69:sum_using_first_and_last_term","label":"(bii) Using First and Last Term Formula","steps":[{"type":"text","order":0,"title":"Identify the parameters","content":"To find the total amount of carbon dioxide sequestered up to the target day, we need to sum the terms of our arithmetic series.\n\nFrom the previous parts of the question, we know:\n- The number of days to reach the target is $n = 54$\n- The amount sequestered on the first day is $a = 18$ tonnes\n- The amount sequestered on the target day ($54^{\\text{th}}$ day) is the last term $l = 150.5$ tonnes","highlights":[{"text":"total amount of carbon dioxide sequestered during the cycle will exceed 4,500 tonnes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_n = \\frac{\\htmlData{anchor=n-var}{n}}{2}(\\htmlData{anchor=a-var}{a} + \\htmlData{anchor=l-var}{l})$"},{"latex":"$$S_{54} = \\frac{\\htmlData{anchor=n-val}{\\color{@sky}{54}}}{2}(\\htmlData{anchor=a-val}{\\color{@amber}{18}} + \\htmlData{anchor=l-val}{\\color{@purple}{150.5}})$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"amber","style":"text-color","anchor":"a-val"},{"color":"purple","style":"text-color","anchor":"l-val"}],"connections":[{"to":"n-val","from":"n-var","color":"sky","label":null,"style":"solid"},{"to":"a-val","from":"a-var","color":"amber","label":null,"style":"solid"},{"to":"l-val","from":"l-var","color":"purple","label":null,"style":"solid"}]},"order":1,"title":"Apply sum formula","description":"We use the arithmetic series sum formula $S_n = \\frac{n}{2}(a+l)$, where $a$ is the first term and $l$ is the last term."},{"type":"text","order":2,"title":"Calculate the total","content":"Now, we evaluate the sum:\n\n$$\\begin{aligned} S_{54} &= 27 \\times (18 + 150.5) \\\\ S_{54} &= 27 \\times 168.5 \\\\ S_{54} &= 4549.5 \\text{ tonnes} \\end{aligned}$$\n\nThis calculation earns the first two marks (**M1** for setting up the sum and **A1** for the correct value).","calculator":[{"gdc":{"expression":"54/2*(18+150.5)","instructions":"Type `54 / 2 * (18 + 150.5)` on the main calculation screen and press ENTER."},"ti84":{"expression":"54/2*(18+150.5)","instructions":"Type `54 / 2 * (18 + 150.5)` on the home screen and press ENTER."},"desmos":{"expression":"\\frac{54}{2}(18 + 150.5)","instructions":"Type `54/2 * (18 + 150.5)` into the expression box to see the result."},"description":"Calculate the total sum directly using basic arithmetic."}]},{"type":"text","order":3,"title":"Make the conclusion","content":"Finally, we compare our calculated sum to the threshold given in the analyst's prediction:\n\n$$4549.5 \\text{ tonnes} > 4500 \\text{ tonnes}$$\n\nSince $4549.5 > 4500$, the analyst's prediction is **correct** (**C1**).","highlights":[{"text":"Determine, with a calculation, whether the analyst's prediction is correct.","location":"specification"}]}]},{"id":"b03c3998-32e7-4f0f-998a-93338f213e69:sum_using_first_term_and_difference","label":"(bii) Using First Term and Common Difference Formula","steps":[{"type":"text","order":0,"title":"Identify key parameters","content":"From the problem description, the daily carbon dioxide capture forms an arithmetic sequence with:\n* First term $a = 18$ tonnes (the amount captured on the first day)\n* Common difference $d = 2.5$ tonnes (the daily increase)\n* Number of days $n = 54$\n\nWe need to find the total amount of carbon dioxide sequestered over these $54$ days and see if it exceeds $4,500$ tonnes.","highlights":[{"text":"captures 18 tonnes of carbon dioxide","location":"specification"},{"text":"increases by 2.5 tonnes","location":"specification"},{"text":"will exceed 4,500 tonnes","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$S_n = \\frac{\\htmlData{anchor=n-val}{\\color{@sky}{n}}}{2}\\left(2\\htmlData{anchor=a-val}{\\color{@amber}{a}} + (\\htmlData{anchor=n-sub}{\\color{@sky}{n}} - 1)\\htmlData{anchor=d-val}{\\color{@purple}{d}}\\right)$"},{"latex":"$$S_{54} = \\frac{\\color{@sky}{54}}{2}\\left(2(\\color{@amber}{18}) + (\\color{@sky}{54} - 1)(\\color{@purple}{2.5})\\right)$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"amber","style":"text-color","anchor":"a-val"},{"color":"purple","style":"text-color","anchor":"d-val"}],"connections":[{"to":"n-sub","from":"n-val","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Apply arithmetic sum formula","description":"We will use the sum formula of an arithmetic series $S_n = \\frac{n}{2}(2a + (n-1)d)$ to compute the total captured dioxide by substituting our values: $a = 18$, $d = 2.5$, and $n = 54$."},{"type":"text","order":2,"title":"Calculate total sum","content":"Let's calculate the value step-by-step:\n\n$$\\begin{aligned} S_{54} &= 27 \\times (36 + 53 \\times 2.5) \\\\ &= 27 \\times (36 + 132.5) \\\\ &= 27 \\times 168.5 \\\\ &= 4549.5 \\text{ tonnes} \\end{aligned}$$\n\nThis calculation earns the first **M1** mark for setting up the sum equation and the **A1** mark for correctly evaluating the sum to $4549.5$.","calculator":[{"gdc":{"expression":"54/2 * (2 * 18 + 53 * 2.5)","instructions":"Enter `54/2 * (2 * 18 + 53 * 2.5)` into the main calculator screen."},"ti84":{"expression":"54/2 * (2 * 18 + 53 * 2.5)","instructions":"Type `54/2 * (2 * 18 + 53 * 2.5)` and press ENTER."},"desmos":{"expression":"54/2 * (2 * 18 + (54 - 1) * 2.5)","instructions":"Type the equation in a single cell: `54/2 * (2 * 18 + (54 - 1) * 2.5)`."},"description":"To calculate the sum on a graphing calculator, simply evaluate the expression directly."}]},{"type":"text","order":3,"title":"Compare and conclude","content":"We compare our calculated value with the analyst's prediction:\n\n$$4549.5 \\text{ tonnes} > 4500 \\text{ tonnes}$$\n\nSince the total amount of carbon dioxide sequestered ($4549.5$ tonnes) is greater than $4500$ tonnes, we conclude that **the analyst's prediction is correct**.\n\nThis final conclusion earns the **C1** mark.","highlights":[{"text":"exceed 4,500 tonnes","location":"specification"}]}]}],"generatedAt":"2026-06-09T06:00:02.607Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"9aadfd8a-f1df-4d17-a73d-c82be4a32f66","revisionId":"91883728-ba11-46b5-a1e6-1125d2357938","questionNumber":null,"questionType":"LA","specification":"The arithmetic series \$S\$ is given by\n\n\\[S = 7 + 11 + 15 + 19 + \\cdots + 303\\]\n\nFind","difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"21372bfe-f35c-4952-b631-215dba2e31d3","label":"a","content":"the number of terms in the series,","markscheme":"- Identifies the first term $a = 7$, common difference $d = 4$ and last term $l = 303$ B1\n- Uses the formula $u_n = a + (n - 1)d$ to set up the equation $303 = 7 + (n - 1)4$ M1\n- Solves the equation to find $n = 75$ A1","marks":3,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"formulaic","label":"Using the n-th Term Formula","steps":[{"type":"text","order":0,"title":"Identify key parameters","content":"To find the number of terms in the arithmetic series, we first need to identify the key components of the sequence:\n\n* **First term ($a$):** The first number in the series is $7$.\n* **Common difference ($d$):** The difference between consecutive terms is $11 - 7 = 4$.\n* **Last term ($u_n$ or $l$):** The final term of the series is $303$.\n\nIdentifying these values correctly earns the first mark B1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_n = \\htmlData{anchor=a-term}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-diff}{\\color{@amber}{d}}$"},{"latex":"$$\\htmlData{anchor=un-term}{\\color{@purple}{303}} = \\htmlData{anchor=a-sub}{\\color{@sky}{7}} + (n - 1)\\htmlData{anchor=d-sub}{\\color{@amber}{4}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-term"},{"color":"amber","style":"text-color","anchor":"d-diff"},{"color":"purple","style":"text-color","anchor":"un-term"}],"connections":[{"to":"a-sub","from":"a-term","color":"sky","label":"first term","style":"solid"},{"to":"d-sub","from":"d-diff","color":"amber","label":"common difference","style":"solid"}]},"order":1,"title":"Set up the equation","description":"We substitute our identified values into the $n$-th term formula for an arithmetic progression. This substitution earns a method mark M1."},{"type":"text","order":2,"title":"Solve for n","content":"Now we solve the equation for $n$ step-by-step:\n\n$$303 = 7 + 4(n - 1)$$\n\n$$296 = 4(n - 1)$$\n\n$$74 = n - 1$$\n\n$$n = 75$$\n\nThus, there are $75$ terms in the series. Finding this value correctly earns the final mark A1.","calculator":[{"gdc":{"expression":"303 = 7 + (X - 1) * 4","instructions":"Go to the Equation menu, select Solver, enter the equation `303 = 7 + (X - 1) * 4`, and solve for $X$."},"ti84":{"expression":"303 = 7 + (N - 1) * 4","instructions":"Press [MATH], scroll down to [Numeric Solver...]. Enter `E1: 303` and `E2: 7 + (N - 1) * 4`, then press [OK] (graph button) and [SOLVE] to find $N$."},"desmos":{"expression":"303 = 7 + (x - 1) * 4","instructions":"Type the equation in the input bar to find the vertical line representing the root."},"description":"You can verify this result using your graphing calculator's equation solver or by graphing."}]},{"type":"text","order":3,"title":"Calculate the sum","content":"Now we find the sum of the series using the sum formula:\n\n$$S_n = \\frac{n}{2}(a + l)$$\n\nSubstituting $n = 75$, $a = 7$, and $l = 303$ into the formula:\n\n$$S_{75} = \\frac{75}{2}(7 + 303)$$\n\n$$S_{75} = \\frac{75}{2}(310)$$\n\n$$S_{75} = 11\\,625$$\n\nThis substitution earns the method marks M1 M1 and the final accuracy mark A1."}]},{"id":"intuitive","label":"Step-Counting Approach","steps":[{"type":"text","order":0,"title":"Identify key terms","content":"First, let's identify the first term, the last term, and the common difference from the given arithmetic series:\n\n$$S = 7 + 11 + 15 + 19 + \\cdots + 303$$\n\n* **First term ($a$):** $7$\n* **Last term ($l$):** $303$\n* **Common difference ($d$):** $11 - 7 = 4$\n\n*This earns you your first mark **[B1]**.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Total Difference} = \\htmlData{anchor=last-term}{\\color{@sky}{303}} - \\htmlData{anchor=first-term}{\\color{@amber}{7}} = 296$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"last-term"},{"color":"amber","style":"text-color","anchor":"first-term"}],"connections":[]},"order":1,"title":"Find the total difference","description":"We calculate the total difference between the last term and the first term to see how much the sequence grows overall."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Number of Steps} = \\frac{\\htmlData{anchor=total-diff}{\\color{@teal}{296}}}{\\htmlData{anchor=common-diff}{\\color{@purple}{4}}} = 74$"}],"highlights":[{"color":"teal","style":"text-color","anchor":"total-diff"},{"color":"purple","style":"text-color","anchor":"common-diff"}],"connections":[]},"order":2,"title":"Calculate the steps","description":"Next, divide the total difference by the common difference to find how many steps (or jumps of $4$) are taken to get from the first term to the last term."},{"type":"text","order":3,"title":"Determine total terms","content":"Since we started at the first term, the number of terms is always **one more** than the number of steps:\n\n$$\\text{Number of terms } (n) = \\text{Number of steps} + 1$$\n\n$$n = 74 + 1$$\n\n$$n = 75$$\n\nThis calculation corresponds to solving the standard formula $l = a + (n - 1)d$, which earns the method mark **[M1]** and the final accuracy mark **[A1]**.\n\nTherefore, there are **$75$** terms in the series.","calculator":[{"gdc":{"expression":"(303-7)/4+1","instructions":"Type (303 - 7) / 4 + 1 and press ENTER."},"ti84":{"expression":"(303-7)/4+1","instructions":"Type (303 - 7) / 4 + 1 and press ENTER."},"desmos":{"expression":"\\frac{303-7}{4}+1","instructions":"Enter the calculation into the input field to get the result."},"description":"You can quickly verify the arithmetic on your calculator."}]},{"type":"text","order":4,"title":"Calculate the sum","content":"With the number of terms known, we can calculate the total sum using the alternative arithmetic sum formula:\n\n$$S_n = \\frac{n}{2}[2a + (n - 1)d]$$\n\nSubstituting $n = 75$, $a = 7$, and $d = 4$:\n\n$$S_{75} = \\frac{75}{2}[2(7) + (75 - 1)4]$$\n\n$$S_{75} = \\frac{75}{2}[14 + 74 \\times 4]$$\n\n$$S_{75} = \\frac{75}{2}[310]$$\n\n$$S_{75} = 11\\,625$$\n\nThis substitution earns **[M1]** and the final sum of **$11\\,625$** earns **[A1]**."}]}],"generatedAt":"2026-06-09T07:59:13.284Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"36eba81c-718b-4b89-8bde-80a4d824f523","label":"b","content":"the sum of the series.","markscheme":"- Uses the arithmetic series sum formula $S_n = \\frac{n}{2}(a + l)$ or $S_n = \\frac{n}{2}(2a + (n - 1)d)$ M1\n- Substitutes $n = 75$, $a = 7$, and $l = 303$ into the formula to get $S_{75} = \\frac{75}{2}(7 + 303)$ M1\n- Obtains the final sum $11\\,625$ A1","marks":3,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"first_last_term_formula","label":"Using First and Last Term Formula","steps":[{"type":"text","order":0,"title":"Identify series parameters","content":"To sum the arithmetic series $S = 7 + 11 + 15 + 19 + \\dots + 303$, we first identify its key components:\n\n- **First term** ($a$) = $7$\n- **Common difference** ($d$) = $11 - 7 = 4$\n- **Last term** ($l$) = $303$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Use the } n\\text{-th term formula: } \\htmlData{anchor=nth-form}{\\color{@sky}{u_n = a + (n - 1)d}}$"},{"latex":"$$\\htmlData{anchor=sub-eq}{\\color{@sky}{303}} = \\htmlData{anchor=a-sub}{\\color{@purple}{7}} + (n - 1)\\htmlData{anchor=d-sub}{\\color{@amber}{4}}$"},{"latex":"$$296 = 4(n - 1)$"},{"latex":"$$n - 1 = 74 \\implies \\htmlData{anchor=n-val}{\\color{@green}{n = 75}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-eq"},{"color":"purple","style":"text-color","anchor":"a-sub"},{"color":"amber","style":"text-color","anchor":"d-sub"},{"color":"green","style":"box","anchor":"n-val"}],"connections":[{"to":"sub-eq","from":"nth-form","color":"sky","label":"Substitute values","style":"solid"}]},"order":1,"title":"Find the number of terms","description":"We need to determine the total number of terms, $n$, in the series. We use the formula for the $n$-th term of an arithmetic progression, $u_n = a + (n - 1)d$, setting $u_n = 303$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Arithmetic series sum formula: } \\htmlData{anchor=sum-formula}{\\color{@teal}{S_n = \\frac{n}{2}(a + l)}}$"},{"latex":"$$\\htmlData{anchor=sum-sub}{S_{75}} = \\frac{\\htmlData{anchor=n-sub}{\\color{@green}{75}}}{2}\\left(\\htmlData{anchor=a-val}{\\color{@purple}{7}} + \\htmlData{anchor=l-val}{\\color{@sky}{303}}\\right)$"},{"latex":"$$S_{75} = 37.5 \\times 310$"},{"latex":"$$S_{75} = \\htmlData{anchor=final-ans}{\\color{@orange}{11\\,625}}$"}],"highlights":[{"color":"green","style":"text-color","anchor":"n-sub"},{"color":"purple","style":"text-color","anchor":"a-val"},{"color":"sky","style":"text-color","anchor":"l-val"},{"color":"orange","style":"box","anchor":"final-ans"}],"connections":[{"to":"sum-sub","from":"sum-formula","color":"teal","label":"Substitute n, a, l","style":"solid"}]},"order":2,"title":"Apply sum formula","description":"Now that we know there are $75$ terms, we can find the sum using the formula $S_n = \\frac{n}{2}(a + l)$, where $l$ is the last term."},{"type":"text","order":3,"title":"Verify with GDC","content":"You can verify this result by evaluating the sum directly using sigma notation on your calculator.","calculator":[{"gdc":{"expression":"\\sum_{X=1}^{75} (7 + (X-1) \\cdot 4)","instructions":"Go to the Run-Matrix menu, press MATH, then Σ(. Input variable X, from 1 to 75, and the expression 7 + (X - 1) × 4."},"ti84":{"expression":"\\sum_{X=1}^{75} (7 + (X-1) \\cdot 4)","instructions":"Press ALPHA then WINDOW (F2), select 2: summation Σ(. Enter the variable X, starting value 1, end value 75, and the formula 7 + (X - 1)*4."},"desmos":{"expression":"\\sum_{n=1}^{75} (7 + (n-1) \\cdot 4)","instructions":"Type sum in an empty expression bar to generate a summation template. Set the limits from n=1 to 75 and input the expression 7 + (n-1)*4."},"description":"Calculate the sum of the series using the sum template."}]}]},{"id":"common_difference_formula","label":"Using Common Difference Formula","steps":[{"type":"text","order":0,"title":"Identify series parameters","content":"From the given arithmetic series:\n\n$$S = 7 + 11 + 15 + 19 + \\dots + 303$$\n\nWe can identify the key parameters:\n\n- First term ($a$) = $7$\n- Common difference ($d$) = $11 - 7 = 4$\n- Last term ($u_n$ or $l$) = $303$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_n = \\htmlData{anchor=a-term}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-diff}{\\color{@amber}{d}}$"},{"latex":"$$303 = \\htmlData{anchor=a-sub}{\\color{@sky}{7}} + (n - 1)\\htmlData{anchor=d-sub}{\\color{@amber}{4}}$"},{"latex":"$$296 = 4(n - 1) \\implies n - 1 = 74 \\implies n = 75$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-term"},{"color":"amber","style":"text-color","anchor":"d-diff"}],"connections":[{"to":"a-sub","from":"a-term","color":"sky","label":"substitute a = 7","style":"solid"},{"to":"d-sub","from":"d-diff","color":"amber","label":"substitute d = 4","style":"solid"}]},"order":1,"title":"Find the number of terms","description":"We use the general $n^{\\text{th}}$ term formula for an arithmetic sequence to solve for $n$, the total number of terms in the series."},{"type":"text","order":2,"title":"Calculate the sum","content":"Now we substitute our values ($n = 75$, $a = 7$, and $d = 4$) into the arithmetic series sum formula involving the common difference:\n\n$$S_n = \\frac{n}{2}\\left(2a + (n - 1)d\\right)$$\n\n$$\\begin{aligned} S_{75} &= \\frac{75}{2}\\left(2(7) + (75 - 1)4\\right) \\\\ &= \\frac{75}{2}\\left(14 + (74)4\\right) \\\\ &= \\frac{75}{2}\\left(14 + 296\\right) \\\\ &= \\frac{75}{2}(310) \\\\ &= 75 \\times 155 \\\\ &= 11\\,625 \\end{aligned}$$\n\nThis earns the method marks M1 M1 and accuracy mark A1.","calculator":[{"gdc":{"expression":"75 / 2 * (2 * 7 + 74 * 4)","instructions":"Enter '75 / 2 * (2 * 7 + 74 * 4)' on the run screen and execute."},"ti84":{"expression":"75 / 2 * (2 * 7 + 74 * 4)","instructions":"Type the arithmetic expression on the home screen: '75 / 2 * (2 * 7 + 74 * 4)' and press ENTER."},"desmos":{"expression":"\\frac{75}{2}\\left(2\\cdot7+\\left(75-1\\right)\\cdot4\\right)","instructions":"Type the formula directly to evaluate the sum."},"description":"Verify the calculation on your graphing calculator."}]}]},{"id":"calculator_sigma_notation","label":"Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Analyze the series","content":"The given arithmetic series is:\n\n$$S = 7 + 11 + 15 + 19 + \\dots + 303$$\n\nTo represent this series in sigma notation, we first find the general term ($r$-th term), $u_r$.\n\n- The first term is $a = 7$.\n- The common difference is $d = 11 - 7 = 4$.\n\nThe formula for the $r$-th term of an arithmetic sequence is:\n\n$$u_r = a + (r - 1)d$$\n\nSubstituting our values in:\n\n$$u_r = 7 + (r - 1)4 = 4r + 3$$"},{"type":"text","order":1,"title":"Find number of terms","content":"Next, we find the number of terms, $n$, by setting the general term equal to the last term, $303$:\n\n$$4n + 3 = 303$$\n\nSolving for $n$:\n\n$$\\begin{aligned} 4n &= 300 \\\\ n &= 75 \\end{aligned}$$\n\nSo, there are $75$ terms in this series."},{"type":"text","order":2,"title":"Write in sigma notation","content":"We can now write the sum of the series using sigma notation as:\n\n$$S = \\sum_{r=1}^{75} (4r + 3)$$\n\nWe can enter this expression directly into a graphing calculator to find the final sum."},{"type":"text","order":3,"title":"Evaluate with calculator","content":"Inputting the summation into your graphing calculator will evaluate the total sum of the series directly.","calculator":[{"gdc":{"expression":"\\sum_{X=1}^{75} (4X+3)","instructions":"Go to the Run-Matrix menu. Press [MATH] then [F6] to see more options, and select Sigma ([F2]).\nFill in the template:\n- Enter the expression: 4X+3\n- Set the variable: X\n- Set the start value: 1\n- Set the end value: 75\nPress [EXE] to calculate."},"ti84":{"expression":"sum(seq(4X+3,X,1,75))","instructions":"Press [ALPHA] [WINDOW] and select 2:summation(\nOr press [MATH] and scroll down to 0:summation(\nFill in the template:\n- Under the Sigma symbol, enter X=1\n- On top of the Sigma symbol, enter 75\n- In the parentheses, enter 4X+3\nPress [ENTER] to calculate."},"desmos":{"expression":"\\sum_{n=1}^{75} (4n+3)","instructions":"Type 'sum' in an expression line to bring up the summation notation.\nSet the lower limit to n=1, the upper limit to 75, and enter 4n+3 inside the summation."},"description":"Evaluate the sum of 4r + 3 from r = 1 to 75."}]},{"type":"text","order":4,"title":"State final answer","content":"The graphing calculator outputs the final sum:\n\n$$S = 11\\,625$$\n\nThis confirms the analytical calculation of the arithmetic series sum."}]}],"generatedAt":"2026-06-09T08:00:56.446Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"21372bfe-f35c-4952-b631-215dba2e31d3:formulaic","label":"(a) Using the n-th Term Formula","steps":[{"type":"text","order":0,"title":"Identify key parameters","content":"To find the number of terms in the arithmetic series, we first need to identify the key components of the sequence:\n\n* **First term ($a$):** The first number in the series is $7$.\n* **Common difference ($d$):** The difference between consecutive terms is $11 - 7 = 4$.\n* **Last term ($u_n$ or $l$):** The final term of the series is $303$.\n\nIdentifying these values correctly earns the first mark B1."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_n = \\htmlData{anchor=a-term}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-diff}{\\color{@amber}{d}}$"},{"latex":"$$\\htmlData{anchor=un-term}{\\color{@purple}{303}} = \\htmlData{anchor=a-sub}{\\color{@sky}{7}} + (n - 1)\\htmlData{anchor=d-sub}{\\color{@amber}{4}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-term"},{"color":"amber","style":"text-color","anchor":"d-diff"},{"color":"purple","style":"text-color","anchor":"un-term"}],"connections":[{"to":"a-sub","from":"a-term","color":"sky","label":"first term","style":"solid"},{"to":"d-sub","from":"d-diff","color":"amber","label":"common difference","style":"solid"}]},"order":1,"title":"Set up the equation","description":"We substitute our identified values into the $n$-th term formula for an arithmetic progression. This substitution earns a method mark M1."},{"type":"text","order":2,"title":"Solve for n","content":"Now we solve the equation for $n$ step-by-step:\n\n$$303 = 7 + 4(n - 1)$$\n\n$$296 = 4(n - 1)$$\n\n$$74 = n - 1$$\n\n$$n = 75$$\n\nThus, there are $75$ terms in the series. Finding this value correctly earns the final mark A1.","calculator":[{"gdc":{"expression":"303 = 7 + (X - 1) * 4","instructions":"Go to the Equation menu, select Solver, enter the equation `303 = 7 + (X - 1) * 4`, and solve for $X$."},"ti84":{"expression":"303 = 7 + (N - 1) * 4","instructions":"Press [MATH], scroll down to [Numeric Solver...]. Enter `E1: 303` and `E2: 7 + (N - 1) * 4`, then press [OK] (graph button) and [SOLVE] to find $N$."},"desmos":{"expression":"303 = 7 + (x - 1) * 4","instructions":"Type the equation in the input bar to find the vertical line representing the root."},"description":"You can verify this result using your graphing calculator's equation solver or by graphing."}]},{"type":"text","order":3,"title":"Calculate the sum","content":"Now we find the sum of the series using the sum formula:\n\n$$S_n = \\frac{n}{2}(a + l)$$\n\nSubstituting $n = 75$, $a = 7$, and $l = 303$ into the formula:\n\n$$S_{75} = \\frac{75}{2}(7 + 303)$$\n\n$$S_{75} = \\frac{75}{2}(310)$$\n\n$$S_{75} = 11\\,625$$\n\nThis substitution earns the method marks M1 M1 and the final accuracy mark A1."}]},{"id":"21372bfe-f35c-4952-b631-215dba2e31d3:intuitive","label":"(a) Step-Counting Approach","steps":[{"type":"text","order":0,"title":"Identify key terms","content":"First, let's identify the first term, the last term, and the common difference from the given arithmetic series:\n\n$$S = 7 + 11 + 15 + 19 + \\cdots + 303$$\n\n* **First term ($a$):** $7$\n* **Last term ($l$):** $303$\n* **Common difference ($d$):** $11 - 7 = 4$\n\n*This earns you your first mark **[B1]**.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Total Difference} = \\htmlData{anchor=last-term}{\\color{@sky}{303}} - \\htmlData{anchor=first-term}{\\color{@amber}{7}} = 296$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"last-term"},{"color":"amber","style":"text-color","anchor":"first-term"}],"connections":[]},"order":1,"title":"Find the total difference","description":"We calculate the total difference between the last term and the first term to see how much the sequence grows overall."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Number of Steps} = \\frac{\\htmlData{anchor=total-diff}{\\color{@teal}{296}}}{\\htmlData{anchor=common-diff}{\\color{@purple}{4}}} = 74$"}],"highlights":[{"color":"teal","style":"text-color","anchor":"total-diff"},{"color":"purple","style":"text-color","anchor":"common-diff"}],"connections":[]},"order":2,"title":"Calculate the steps","description":"Next, divide the total difference by the common difference to find how many steps (or jumps of $4$) are taken to get from the first term to the last term."},{"type":"text","order":3,"title":"Determine total terms","content":"Since we started at the first term, the number of terms is always **one more** than the number of steps:\n\n$$\\text{Number of terms } (n) = \\text{Number of steps} + 1$$\n\n$$n = 74 + 1$$\n\n$$n = 75$$\n\nThis calculation corresponds to solving the standard formula $l = a + (n - 1)d$, which earns the method mark **[M1]** and the final accuracy mark **[A1]**.\n\nTherefore, there are **$75$** terms in the series.","calculator":[{"gdc":{"expression":"(303-7)/4+1","instructions":"Type (303 - 7) / 4 + 1 and press ENTER."},"ti84":{"expression":"(303-7)/4+1","instructions":"Type (303 - 7) / 4 + 1 and press ENTER."},"desmos":{"expression":"\\frac{303-7}{4}+1","instructions":"Enter the calculation into the input field to get the result."},"description":"You can quickly verify the arithmetic on your calculator."}]},{"type":"text","order":4,"title":"Calculate the sum","content":"With the number of terms known, we can calculate the total sum using the alternative arithmetic sum formula:\n\n$$S_n = \\frac{n}{2}[2a + (n - 1)d]$$\n\nSubstituting $n = 75$, $a = 7$, and $d = 4$:\n\n$$S_{75} = \\frac{75}{2}[2(7) + (75 - 1)4]$$\n\n$$S_{75} = \\frac{75}{2}[14 + 74 \\times 4]$$\n\n$$S_{75} = \\frac{75}{2}[310]$$\n\n$$S_{75} = 11\\,625$$\n\nThis substitution earns **[M1]** and the final sum of **$11\\,625$** earns **[A1]**."}]},{"id":"36eba81c-718b-4b89-8bde-80a4d824f523:first_last_term_formula","label":"(b) Using First and Last Term Formula","steps":[{"type":"text","order":0,"title":"Identify series parameters","content":"To sum the arithmetic series $S = 7 + 11 + 15 + 19 + \\dots + 303$, we first identify its key components:\n\n- **First term** ($a$) = $7$\n- **Common difference** ($d$) = $11 - 7 = 4$\n- **Last term** ($l$) = $303$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Use the } n\\text{-th term formula: } \\htmlData{anchor=nth-form}{\\color{@sky}{u_n = a + (n - 1)d}}$"},{"latex":"$$\\htmlData{anchor=sub-eq}{\\color{@sky}{303}} = \\htmlData{anchor=a-sub}{\\color{@purple}{7}} + (n - 1)\\htmlData{anchor=d-sub}{\\color{@amber}{4}}$"},{"latex":"$$296 = 4(n - 1)$"},{"latex":"$$n - 1 = 74 \\implies \\htmlData{anchor=n-val}{\\color{@green}{n = 75}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-eq"},{"color":"purple","style":"text-color","anchor":"a-sub"},{"color":"amber","style":"text-color","anchor":"d-sub"},{"color":"green","style":"box","anchor":"n-val"}],"connections":[{"to":"sub-eq","from":"nth-form","color":"sky","label":"Substitute values","style":"solid"}]},"order":1,"title":"Find the number of terms","description":"We need to determine the total number of terms, $n$, in the series. We use the formula for the $n$-th term of an arithmetic progression, $u_n = a + (n - 1)d$, setting $u_n = 303$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Arithmetic series sum formula: } \\htmlData{anchor=sum-formula}{\\color{@teal}{S_n = \\frac{n}{2}(a + l)}}$"},{"latex":"$$\\htmlData{anchor=sum-sub}{S_{75}} = \\frac{\\htmlData{anchor=n-sub}{\\color{@green}{75}}}{2}\\left(\\htmlData{anchor=a-val}{\\color{@purple}{7}} + \\htmlData{anchor=l-val}{\\color{@sky}{303}}\\right)$"},{"latex":"$$S_{75} = 37.5 \\times 310$"},{"latex":"$$S_{75} = \\htmlData{anchor=final-ans}{\\color{@orange}{11\\,625}}$"}],"highlights":[{"color":"green","style":"text-color","anchor":"n-sub"},{"color":"purple","style":"text-color","anchor":"a-val"},{"color":"sky","style":"text-color","anchor":"l-val"},{"color":"orange","style":"box","anchor":"final-ans"}],"connections":[{"to":"sum-sub","from":"sum-formula","color":"teal","label":"Substitute n, a, l","style":"solid"}]},"order":2,"title":"Apply sum formula","description":"Now that we know there are $75$ terms, we can find the sum using the formula $S_n = \\frac{n}{2}(a + l)$, where $l$ is the last term."},{"type":"text","order":3,"title":"Verify with GDC","content":"You can verify this result by evaluating the sum directly using sigma notation on your calculator.","calculator":[{"gdc":{"expression":"\\sum_{X=1}^{75} (7 + (X-1) \\cdot 4)","instructions":"Go to the Run-Matrix menu, press MATH, then Σ(. Input variable X, from 1 to 75, and the expression 7 + (X - 1) × 4."},"ti84":{"expression":"\\sum_{X=1}^{75} (7 + (X-1) \\cdot 4)","instructions":"Press ALPHA then WINDOW (F2), select 2: summation Σ(. Enter the variable X, starting value 1, end value 75, and the formula 7 + (X - 1)*4."},"desmos":{"expression":"\\sum_{n=1}^{75} (7 + (n-1) \\cdot 4)","instructions":"Type sum in an empty expression bar to generate a summation template. Set the limits from n=1 to 75 and input the expression 7 + (n-1)*4."},"description":"Calculate the sum of the series using the sum template."}]}]},{"id":"36eba81c-718b-4b89-8bde-80a4d824f523:common_difference_formula","label":"(b) Using Common Difference Formula","steps":[{"type":"text","order":0,"title":"Identify series parameters","content":"From the given arithmetic series:\n\n$$S = 7 + 11 + 15 + 19 + \\dots + 303$$\n\nWe can identify the key parameters:\n\n- First term ($a$) = $7$\n- Common difference ($d$) = $11 - 7 = 4$\n- Last term ($u_n$ or $l$) = $303$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_n = \\htmlData{anchor=a-term}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-diff}{\\color{@amber}{d}}$"},{"latex":"$$303 = \\htmlData{anchor=a-sub}{\\color{@sky}{7}} + (n - 1)\\htmlData{anchor=d-sub}{\\color{@amber}{4}}$"},{"latex":"$$296 = 4(n - 1) \\implies n - 1 = 74 \\implies n = 75$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-term"},{"color":"amber","style":"text-color","anchor":"d-diff"}],"connections":[{"to":"a-sub","from":"a-term","color":"sky","label":"substitute a = 7","style":"solid"},{"to":"d-sub","from":"d-diff","color":"amber","label":"substitute d = 4","style":"solid"}]},"order":1,"title":"Find the number of terms","description":"We use the general $n^{\\text{th}}$ term formula for an arithmetic sequence to solve for $n$, the total number of terms in the series."},{"type":"text","order":2,"title":"Calculate the sum","content":"Now we substitute our values ($n = 75$, $a = 7$, and $d = 4$) into the arithmetic series sum formula involving the common difference:\n\n$$S_n = \\frac{n}{2}\\left(2a + (n - 1)d\\right)$$\n\n$$\\begin{aligned} S_{75} &= \\frac{75}{2}\\left(2(7) + (75 - 1)4\\right) \\\\ &= \\frac{75}{2}\\left(14 + (74)4\\right) \\\\ &= \\frac{75}{2}\\left(14 + 296\\right) \\\\ &= \\frac{75}{2}(310) \\\\ &= 75 \\times 155 \\\\ &= 11\\,625 \\end{aligned}$$\n\nThis earns the method marks M1 M1 and accuracy mark A1.","calculator":[{"gdc":{"expression":"75 / 2 * (2 * 7 + 74 * 4)","instructions":"Enter '75 / 2 * (2 * 7 + 74 * 4)' on the run screen and execute."},"ti84":{"expression":"75 / 2 * (2 * 7 + 74 * 4)","instructions":"Type the arithmetic expression on the home screen: '75 / 2 * (2 * 7 + 74 * 4)' and press ENTER."},"desmos":{"expression":"\\frac{75}{2}\\left(2\\cdot7+\\left(75-1\\right)\\cdot4\\right)","instructions":"Type the formula directly to evaluate the sum."},"description":"Verify the calculation on your graphing calculator."}]}]},{"id":"36eba81c-718b-4b89-8bde-80a4d824f523:calculator_sigma_notation","label":"(b) Using Graphing Calculator","steps":[{"type":"text","order":0,"title":"Analyze the series","content":"The given arithmetic series is:\n\n$$S = 7 + 11 + 15 + 19 + \\dots + 303$$\n\nTo represent this series in sigma notation, we first find the general term ($r$-th term), $u_r$.\n\n- The first term is $a = 7$.\n- The common difference is $d = 11 - 7 = 4$.\n\nThe formula for the $r$-th term of an arithmetic sequence is:\n\n$$u_r = a + (r - 1)d$$\n\nSubstituting our values in:\n\n$$u_r = 7 + (r - 1)4 = 4r + 3$$"},{"type":"text","order":1,"title":"Find number of terms","content":"Next, we find the number of terms, $n$, by setting the general term equal to the last term, $303$:\n\n$$4n + 3 = 303$$\n\nSolving for $n$:\n\n$$\\begin{aligned} 4n &= 300 \\\\ n &= 75 \\end{aligned}$$\n\nSo, there are $75$ terms in this series."},{"type":"text","order":2,"title":"Write in sigma notation","content":"We can now write the sum of the series using sigma notation as:\n\n$$S = \\sum_{r=1}^{75} (4r + 3)$$\n\nWe can enter this expression directly into a graphing calculator to find the final sum."},{"type":"text","order":3,"title":"Evaluate with calculator","content":"Inputting the summation into your graphing calculator will evaluate the total sum of the series directly.","calculator":[{"gdc":{"expression":"\\sum_{X=1}^{75} (4X+3)","instructions":"Go to the Run-Matrix menu. Press [MATH] then [F6] to see more options, and select Sigma ([F2]).\nFill in the template:\n- Enter the expression: 4X+3\n- Set the variable: X\n- Set the start value: 1\n- Set the end value: 75\nPress [EXE] to calculate."},"ti84":{"expression":"sum(seq(4X+3,X,1,75))","instructions":"Press [ALPHA] [WINDOW] and select 2:summation(\nOr press [MATH] and scroll down to 0:summation(\nFill in the template:\n- Under the Sigma symbol, enter X=1\n- On top of the Sigma symbol, enter 75\n- In the parentheses, enter 4X+3\nPress [ENTER] to calculate."},"desmos":{"expression":"\\sum_{n=1}^{75} (4n+3)","instructions":"Type 'sum' in an expression line to bring up the summation notation.\nSet the lower limit to n=1, the upper limit to 75, and enter 4n+3 inside the summation."},"description":"Evaluate the sum of 4r + 3 from r = 1 to 75."}]},{"type":"text","order":4,"title":"State final answer","content":"The graphing calculator outputs the final sum:\n\n$$S = 11\\,625$$\n\nThis confirms the analytical calculation of the arithmetic series sum."}]}],"generatedAt":"2026-06-09T08:01:10.012Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"c5160e74-6e5c-4d5f-8399-60a97f2a00d0","revisionId":"a5556d05-bdac-41d7-ad01-19ccce29715b","questionNumber":null,"questionType":"LA","specification":"Liam is saving money to purchase a high-end camera. He saves 6.00 in week 1, 6.40 in week 2, 6.80 in week 3 and so on, continuing this pattern until he has enough total savings to buy the camera.\n\nHe decides to model his savings using either an arithmetic series or a geometric series. Using the information given,","difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"5cedbaab-672f-422c-911e-d5ab2b8de40e","label":"a","content":"(i) state with a reason whether an arithmetic series or a geometric series should be used,\n\n(ii) write down an expression, in terms of \$n\$, for the amount, in pounds ( ), saved in week \$n\$.","markscheme":"- Identifies arithmetic series because there is a constant common difference of $0.40$ between successive weeks (e.g. $6.40 - 6.00 = 0.40$) B1\n- States the expression $u_n = 6 + 0.4(n - 1)$ or $u_n = 5.6 + 0.4n$ B1","marks":2,"order":0,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"direct_analysis","label":"Direct Analysis","steps":[{"type":"text","order":0,"title":"Analyze the pattern","content":"To determine whether Liam's savings follow an arithmetic or a geometric series, let's look at the amounts saved in the first few weeks:\n- Week 1: $\\pounds 6.00$\n- Week 2: $\\pounds 6.40$\n- Week 3: $\\pounds 6.80$\n\nLet's calculate the difference between consecutive weeks:\n- From Week 1 to Week 2: \n$$\\pounds 6.40 - \\pounds 6.00 = \\pounds 0.40$$\n- From Week 2 to Week 3: \n$$\\pounds 6.80 - \\pounds 6.40 = \\pounds 0.40$$\n\nSince the savings increase by a constant amount each week, we have a constant common difference.","highlights":[{"text":"6.00 in week 1","location":"specification"},{"text":"6.40 in week 2","location":"specification"},{"text":"6.80 in week 3","location":"specification"}]},{"type":"text","order":1,"title":"State the series type","content":"For Part (i), since there is a constant common difference of $\\pounds 0.40$ between successive weeks, we must model his savings using an **arithmetic series**.\n\n*Award 1 mark (B1) for stating arithmetic series with this reason.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_n = \\htmlData{anchor=a-term}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-term}{\\color{@amber}{d}}$"},{"latex":"$$u_n = \\htmlData{anchor=a-val}{\\color{@sky}{6}} + (n - 1)(\\htmlData{anchor=d-val}{\\color{@amber}{0.4}})$"},{"latex":"$$u_n = 6 + 0.4n - 0.4 = 5.6 + 0.4n$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-term"},{"color":"amber","style":"text-color","anchor":"d-term"}],"connections":[{"to":"a-term","from":"a-val","color":"sky","label":"first term","style":"solid"},{"to":"d-term","from":"d-val","color":"amber","label":"common difference","style":"solid"}]},"order":2,"title":"Apply arithmetic formula","description":"For Part (ii), we use the formula for the $n$-th term of an arithmetic progression: $$u_n = a + (n - 1)d$$ where $a$ is the first term and $d$ is the common difference. Substituting our values $a = 6$ and $d = 0.4$ gives the expression."},{"type":"text","order":3,"title":"State the final expression","content":"The expression for the amount saved in week $n$ can be written as:\n\n$$u_n = 6 + 0.4(n - 1)$$\n\nor simplified to:\n\n$$u_n = 5.6 + 0.4n$$\n\n*Award 1 mark (B1) for writing down either version of this expression.*"}]}],"generatedAt":"2026-06-09T09:22:01.110Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"d17a86e5-bbd5-4a8e-8384-5fe8584dec57","label":"b","content":"Given that the camera Liam wants to buy costs 560, find the number of weeks it will take for Liam to save enough money to buy the camera.","markscheme":"- Uses the sum of an arithmetic series formula $S_n = \\frac{n}{2}[2a + (n - 1)d]$ with $a = 6$ and $d = 0.4$ M1\n- Sets up the equation or inequality $\\frac{n}{2}[12 + 0.4(n - 1)] \\ge 560$ and simplifies to a quadratic such as $n^2 + 29n - 2800 = 0$ M1\n- Solves the quadratic to find the positive root $n \\approx 40.4$ M1\n- Concludes $n = 41$ weeks are needed for Liam to save enough money A1","marks":4,"order":1,"markingRubric":null,"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Method","steps":[{"type":"text","order":0,"title":"Identify arithmetic series parameters","content":"From the problem description, Liam's savings form an arithmetic series where:\n\n* The first term is $a = 6$ (the savings in week 1)\n* The common difference is $d = 0.4$ (each week his savings increase by $0.40$)\n\nWe want to find the number of weeks $n$ required for his total savings, $S_n$, to reach at least $560$:\n\n$$S_n \\ge 560$$","highlights":[{"text":"costs 560","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{The sum of an arithmetic series is given by:}$"},{"latex":"$$$S_n = \\frac{n}{2} \\left[ 2\\htmlData{anchor=a-var}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-var}{\\color{@amber}{d}} \\right]$$"},{"latex":"$$$\\text{Substituting } \\htmlData{anchor=a-val}{\\color{@sky}{a = 6}} \\text{ and } \\htmlData{anchor=d-val}{\\color{@amber}{d = 0.4}} \\text{:}$$"},{"latex":"$$$S_n = \\frac{n}{2} \\left[ 2(\\color{@sky}{6}) + (n - 1)(\\color{@amber}{0.4}) \\right]$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"d-val"}],"connections":[{"to":"a-var","from":"a-val","color":"sky","label":"Substitute a","style":"solid"},{"to":"d-var","from":"d-val","color":"amber","label":"Substitute d","style":"solid"}]},"order":1,"title":"Recall sum formula","description":"We substitute our parameters into the general arithmetic series sum formula."},{"type":"text","order":2,"title":"Formulate quadratic equation","content":"Set the sum expression equal to $560$ to find the boundary value for $n$:\n\n$$\\frac{n}{2} \\left[ 12 + 0.4(n - 1) \\right] = 560$$\n\nMultiply both sides by $2$ to eliminate the fraction:\n\n$$n \\left[ 12 + 0.4n - 0.4 \\right] = 1120$$\n\nSimplify the terms inside the bracket:\n\n$$n \\left[ 11.6 + 0.4n \\right] = 1120$$\n\n$$0.4n^2 + 11.6n - 1120 = 0$$\n\nTo make the coefficients integers, we can multiply the entire equation by $2.5$ (or divide by $0.4$):\n\n$$n^2 + 29n - 2800 = 0$$"},{"type":"text","order":3,"title":"Solve quadratic equation analytically","content":"Using the quadratic formula, $n = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 29$, and $c = -2800$:\n\n$$n = \\frac{-29 \\pm \\sqrt{29^2 - 4(1)(-2800)}}{2(1)}$$\n\n$$n = \\frac{-29 \\pm \\sqrt{841 + 11200}}{2}$$\n\n$$n = \\frac{-29 \\pm \\sqrt{12041}}{2}$$\n\nEvaluating this numerically, we get:\n\n$$n \\approx \\frac{-29 \\pm 109.73}{2}$$\n\nSince $n$ must be positive, we take the positive root:\n\n$$n \\approx \\frac{80.73}{2} \\approx 40.37$$"},{"type":"text","order":4,"title":"Solve using GDC","content":"You can also quickly solve this quadratic using your graphing calculator or software.","calculator":[{"gdc":{"expression":"x^2 + 29x - 2800 = 0","instructions":"Go to the Equation menu, select Polynomial, choose Degree 2, and enter the coefficients $a=1, b=29, c=-2800$ to find the positive root."},"ti84":{"expression":"x^2 + 29x - 2800 = 0","instructions":"Press [APPS] and select 'PlySmlt2'. Choose 'Polynomial Root Finder'. Set Order to 2, click Next, and enter $a_2 = 1$, $a_1 = 29$, and $a_0 = -2800$. Press [SOLVE] to get the roots."},"desmos":{"expression":"y = x^2 + 29x - 2800","instructions":"Type $y = x^2 + 29x - 2800$ in the input line and find the positive x-intercept by clicking on the curve where it crosses the positive x-axis."},"description":"Solve $n^2 + 29n - 2800 = 0$ using the solver tool or by graphing."}]},{"type":"text","order":5,"title":"Round up for conclusion","content":"Since $n \\approx 40.4$ weeks, completing $40$ weeks of saving will not be quite enough to reach $560$.\n\nTherefore, we must round up to the next whole number to ensure his savings meet or exceed the cost of the camera.\n\n$$n = 41 \\text{ weeks}$$"}]},{"id":"gdc","label":"Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Identify sequence parameters","content":"From the question, we know Liam saves $6.00$ in week 1, $6.40$ in week 2, and $6.80$ in week 3. This pattern represents an arithmetic series with the following parameters:\n- First term ($a$) = $6$\n- Common difference ($d$) = $0.4$\n\nTo find when he can purchase the camera, we need his total savings ($S_n$) to be at least $560$."},{"type":"text","order":1,"title":"Set up sum formula","content":"The sum of the first $n$ terms of an arithmetic series is given by the formula from your reference booklet:\n\n$$S_n = \\frac{n}{2}[2a + (n - 1)d]$$\n\nSubstituting $a = 6$ and $d = 0.4$:\n\n$$\\begin{aligned} S_n &= \\frac{n}{2}[2(6) + 0.4(n - 1)] \\\\ S_n &= \\frac{n}{2}[12 + 0.4n - 0.4] \\\\ S_n &= \\frac{n}{2}[11.6 + 0.4n] \\\\ S_n &= 5.8n + 0.2n^2 \\end{aligned}$$\n\nWe need to find the smallest integer $n$ such that $S_n \\ge 560$."},{"type":"text","order":2,"title":"Use graphing calculator","content":"Instead of solving the quadratic equation manually, we can use a graphing calculator (GDC) to plot the sum function $y = 0.2x^2 + 5.8x$ and find where it intersects $y = 560$, or inspect a table of values to find the first integer value of $x$ where $y$ exceeds $560$.","calculator":[{"gdc":{"expression":"f1(x) = 0.2*x^2 + 5.8*x \\n f2(x) = 560","instructions":"1. Go to the Graph application.\\n2. Enter the functions f1(x) = 0.2x^2 + 5.8x and f2(x) = 560.\\n3. Adjust the window to show x from 0 to 50 and y from 0 to 600.\\n4. Use the Analyze Graph tool to find the Intersection point.\\n5. The intersection coordinate will show x ≈ 40.4."},"ti84":{"expression":"Y1 = 0.2X^2 + 5.8X \\n Y2 = 560","instructions":"1. Press [Y=] and enter \\nY1 = 0.2X^2 + 5.8X\\nY2 = 560\\n2. Press [WINDOW] and set Xmin = 0, Xmax = 50, Ymin = 0, Ymax = 600.\\n3. Press [GRAPH].\\n4. Press [2nd] [CALC] and select 5: intersect.\\n5. Press [ENTER] three times to find the intersection point: X ≈ 40.4."},"desmos":{"expression":"y = 0.2x^2 + 5.8x \\n y = 560","instructions":"1. Type y = 0.2x^2 + 5.8x into the first row.\\n2. Type y = 560 into the second row.\\n3. Click on the point of intersection on the graph to display its coordinates.\\n4. The intersection point is (40.404, 560)."},"description":"Find the intersection of $y = 0.2x^2 + 5.8x$ and $y = 560$ to determine when total savings reach $560$."}]},{"type":"text","order":3,"title":"Interpret the result","content":"The intersection point is at $n \\approx 40.4$ weeks. \n\nBecause savings are added at the end of each week, we need the first whole number of weeks that exceeds $40.4$.\n- At $n = 40$ weeks, Liam's savings will be:\n $$S_{40} = 0.2(40)^2 + 5.8(40) = 320 + 232 = 552$$\n This is not enough ($552 < 560$).\n- At $n = 41$ weeks, Liam's savings will be:\n $$S_{41} = 0.2(41)^2 + 5.8(41) = 336.2 + 237.8 = 574$$\n This is enough ($574 \\ge 560$).\n\nTherefore, it will take Liam **$41$ weeks** to save enough money to buy the camera."}]}],"generatedAt":"2026-06-09T09:23:03.785Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"5cedbaab-672f-422c-911e-d5ab2b8de40e:direct_analysis","label":"(a) Direct Analysis","steps":[{"type":"text","order":0,"title":"Analyze the pattern","content":"To determine whether Liam's savings follow an arithmetic or a geometric series, let's look at the amounts saved in the first few weeks:\n- Week 1: $\\pounds 6.00$\n- Week 2: $\\pounds 6.40$\n- Week 3: $\\pounds 6.80$\n\nLet's calculate the difference between consecutive weeks:\n- From Week 1 to Week 2: \n$$\\pounds 6.40 - \\pounds 6.00 = \\pounds 0.40$$\n- From Week 2 to Week 3: \n$$\\pounds 6.80 - \\pounds 6.40 = \\pounds 0.40$$\n\nSince the savings increase by a constant amount each week, we have a constant common difference.","highlights":[{"text":"6.00 in week 1","location":"specification"},{"text":"6.40 in week 2","location":"specification"},{"text":"6.80 in week 3","location":"specification"}]},{"type":"text","order":1,"title":"State the series type","content":"For Part (i), since there is a constant common difference of $\\pounds 0.40$ between successive weeks, we must model his savings using an **arithmetic series**.\n\n*Award 1 mark (B1) for stating arithmetic series with this reason.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u_n = \\htmlData{anchor=a-term}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-term}{\\color{@amber}{d}}$"},{"latex":"$$u_n = \\htmlData{anchor=a-val}{\\color{@sky}{6}} + (n - 1)(\\htmlData{anchor=d-val}{\\color{@amber}{0.4}})$"},{"latex":"$$u_n = 6 + 0.4n - 0.4 = 5.6 + 0.4n$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-term"},{"color":"amber","style":"text-color","anchor":"d-term"}],"connections":[{"to":"a-term","from":"a-val","color":"sky","label":"first term","style":"solid"},{"to":"d-term","from":"d-val","color":"amber","label":"common difference","style":"solid"}]},"order":2,"title":"Apply arithmetic formula","description":"For Part (ii), we use the formula for the $n$-th term of an arithmetic progression: $$u_n = a + (n - 1)d$$ where $a$ is the first term and $d$ is the common difference. Substituting our values $a = 6$ and $d = 0.4$ gives the expression."},{"type":"text","order":3,"title":"State the final expression","content":"The expression for the amount saved in week $n$ can be written as:\n\n$$u_n = 6 + 0.4(n - 1)$$\n\nor simplified to:\n\n$$u_n = 5.6 + 0.4n$$\n\n*Award 1 mark (B1) for writing down either version of this expression.*"}]},{"id":"d17a86e5-bbd5-4a8e-8384-5fe8584dec57:algebraic","label":"(b) Algebraic Method","steps":[{"type":"text","order":0,"title":"Identify arithmetic series parameters","content":"From the problem description, Liam's savings form an arithmetic series where:\n\n* The first term is $a = 6$ (the savings in week 1)\n* The common difference is $d = 0.4$ (each week his savings increase by $0.40$)\n\nWe want to find the number of weeks $n$ required for his total savings, $S_n$, to reach at least $560$:\n\n$$S_n \\ge 560$$","highlights":[{"text":"costs 560","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{The sum of an arithmetic series is given by:}$"},{"latex":"$$$S_n = \\frac{n}{2} \\left[ 2\\htmlData{anchor=a-var}{\\color{@sky}{a}} + (n - 1)\\htmlData{anchor=d-var}{\\color{@amber}{d}} \\right]$$"},{"latex":"$$$\\text{Substituting } \\htmlData{anchor=a-val}{\\color{@sky}{a = 6}} \\text{ and } \\htmlData{anchor=d-val}{\\color{@amber}{d = 0.4}} \\text{:}$$"},{"latex":"$$$S_n = \\frac{n}{2} \\left[ 2(\\color{@sky}{6}) + (n - 1)(\\color{@amber}{0.4}) \\right]$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"d-val"}],"connections":[{"to":"a-var","from":"a-val","color":"sky","label":"Substitute a","style":"solid"},{"to":"d-var","from":"d-val","color":"amber","label":"Substitute d","style":"solid"}]},"order":1,"title":"Recall sum formula","description":"We substitute our parameters into the general arithmetic series sum formula."},{"type":"text","order":2,"title":"Formulate quadratic equation","content":"Set the sum expression equal to $560$ to find the boundary value for $n$:\n\n$$\\frac{n}{2} \\left[ 12 + 0.4(n - 1) \\right] = 560$$\n\nMultiply both sides by $2$ to eliminate the fraction:\n\n$$n \\left[ 12 + 0.4n - 0.4 \\right] = 1120$$\n\nSimplify the terms inside the bracket:\n\n$$n \\left[ 11.6 + 0.4n \\right] = 1120$$\n\n$$0.4n^2 + 11.6n - 1120 = 0$$\n\nTo make the coefficients integers, we can multiply the entire equation by $2.5$ (or divide by $0.4$):\n\n$$n^2 + 29n - 2800 = 0$$"},{"type":"text","order":3,"title":"Solve quadratic equation analytically","content":"Using the quadratic formula, $n = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 29$, and $c = -2800$:\n\n$$n = \\frac{-29 \\pm \\sqrt{29^2 - 4(1)(-2800)}}{2(1)}$$\n\n$$n = \\frac{-29 \\pm \\sqrt{841 + 11200}}{2}$$\n\n$$n = \\frac{-29 \\pm \\sqrt{12041}}{2}$$\n\nEvaluating this numerically, we get:\n\n$$n \\approx \\frac{-29 \\pm 109.73}{2}$$\n\nSince $n$ must be positive, we take the positive root:\n\n$$n \\approx \\frac{80.73}{2} \\approx 40.37$$"},{"type":"text","order":4,"title":"Solve using GDC","content":"You can also quickly solve this quadratic using your graphing calculator or software.","calculator":[{"gdc":{"expression":"x^2 + 29x - 2800 = 0","instructions":"Go to the Equation menu, select Polynomial, choose Degree 2, and enter the coefficients $a=1, b=29, c=-2800$ to find the positive root."},"ti84":{"expression":"x^2 + 29x - 2800 = 0","instructions":"Press [APPS] and select 'PlySmlt2'. Choose 'Polynomial Root Finder'. Set Order to 2, click Next, and enter $a_2 = 1$, $a_1 = 29$, and $a_0 = -2800$. Press [SOLVE] to get the roots."},"desmos":{"expression":"y = x^2 + 29x - 2800","instructions":"Type $y = x^2 + 29x - 2800$ in the input line and find the positive x-intercept by clicking on the curve where it crosses the positive x-axis."},"description":"Solve $n^2 + 29n - 2800 = 0$ using the solver tool or by graphing."}]},{"type":"text","order":5,"title":"Round up for conclusion","content":"Since $n \\approx 40.4$ weeks, completing $40$ weeks of saving will not be quite enough to reach $560$.\n\nTherefore, we must round up to the next whole number to ensure his savings meet or exceed the cost of the camera.\n\n$$n = 41 \\text{ weeks}$$"}]},{"id":"d17a86e5-bbd5-4a8e-8384-5fe8584dec57:gdc","label":"(b) Using a Graphing Calculator","steps":[{"type":"text","order":0,"title":"Identify sequence parameters","content":"From the question, we know Liam saves $6.00$ in week 1, $6.40$ in week 2, and $6.80$ in week 3. This pattern represents an arithmetic series with the following parameters:\n- First term ($a$) = $6$\n- Common difference ($d$) = $0.4$\n\nTo find when he can purchase the camera, we need his total savings ($S_n$) to be at least $560$."},{"type":"text","order":1,"title":"Set up sum formula","content":"The sum of the first $n$ terms of an arithmetic series is given by the formula from your reference booklet:\n\n$$S_n = \\frac{n}{2}[2a + (n - 1)d]$$\n\nSubstituting $a = 6$ and $d = 0.4$:\n\n$$\\begin{aligned} S_n &= \\frac{n}{2}[2(6) + 0.4(n - 1)] \\\\ S_n &= \\frac{n}{2}[12 + 0.4n - 0.4] \\\\ S_n &= \\frac{n}{2}[11.6 + 0.4n] \\\\ S_n &= 5.8n + 0.2n^2 \\end{aligned}$$\n\nWe need to find the smallest integer $n$ such that $S_n \\ge 560$."},{"type":"text","order":2,"title":"Use graphing calculator","content":"Instead of solving the quadratic equation manually, we can use a graphing calculator (GDC) to plot the sum function $y = 0.2x^2 + 5.8x$ and find where it intersects $y = 560$, or inspect a table of values to find the first integer value of $x$ where $y$ exceeds $560$.","calculator":[{"gdc":{"expression":"f1(x) = 0.2*x^2 + 5.8*x \\n f2(x) = 560","instructions":"1. Go to the Graph application.\\n2. Enter the functions f1(x) = 0.2x^2 + 5.8x and f2(x) = 560.\\n3. Adjust the window to show x from 0 to 50 and y from 0 to 600.\\n4. Use the Analyze Graph tool to find the Intersection point.\\n5. The intersection coordinate will show x ≈ 40.4."},"ti84":{"expression":"Y1 = 0.2X^2 + 5.8X \\n Y2 = 560","instructions":"1. Press [Y=] and enter \\nY1 = 0.2X^2 + 5.8X\\nY2 = 560\\n2. Press [WINDOW] and set Xmin = 0, Xmax = 50, Ymin = 0, Ymax = 600.\\n3. Press [GRAPH].\\n4. Press [2nd] [CALC] and select 5: intersect.\\n5. Press [ENTER] three times to find the intersection point: X ≈ 40.4."},"desmos":{"expression":"y = 0.2x^2 + 5.8x \\n y = 560","instructions":"1. Type y = 0.2x^2 + 5.8x into the first row.\\n2. Type y = 560 into the second row.\\n3. Click on the point of intersection on the graph to display its coordinates.\\n4. The intersection point is (40.404, 560)."},"description":"Find the intersection of $y = 0.2x^2 + 5.8x$ and $y = 560$ to determine when total savings reach $560$."}]},{"type":"text","order":3,"title":"Interpret the result","content":"The intersection point is at $n \\approx 40.4$ weeks. \n\nBecause savings are added at the end of each week, we need the first whole number of weeks that exceeds $40.4$.\n- At $n = 40$ weeks, Liam's savings will be:\n $$S_{40} = 0.2(40)^2 + 5.8(40) = 320 + 232 = 552$$\n This is not enough ($552 < 560$).\n- At $n = 41$ weeks, Liam's savings will be:\n $$S_{41} = 0.2(41)^2 + 5.8(41) = 336.2 + 237.8 = 574$$\n This is enough ($574 \\ge 560$).\n\nTherefore, it will take Liam **$41$ weeks** to save enough money to buy the camera."}]}],"generatedAt":"2026-06-09T09:23:15.225Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"2dbce232-c0d7-4dd6-b139-3eeb4b6a2e55","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"5b95689d-71eb-476e-a8e6-a38dbb8ea544","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"842a134f-0c6b-4eca-b37d-70f1a75095c1","difficultyBand":"easy","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"02e9df4b-1c3b-4086-89c5-ff1e95c0f4d4","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"a2e05815-84d8-4df1-b228-38d7561163b2","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"eb9eaff0-b538-4f88-99de-ae22ee7a57e7","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"626e5417-01b1-4503-83ee-86627c92915f","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"aff5e867-ea3b-4a26-9a7c-0af68ae1a615","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"ef9f3562-22be-429e-a8e5-272e03765a58","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"2d180c3d-5e91-4921-b7d0-df95ec1d9dcd","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"2e6bfee7-d832-4871-992e-3f03f37d8da2","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"3c05112e-6358-4470-a53f-1e101c2a2266","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"4139864f-c551-437f-be46-9222f2533947","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"414a7b99-1f37-4b28-b01c-52b671747e8d","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"4385428a-c8d1-4c62-a40a-6fe26604b610","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"4e52a92e-73be-4fae-8521-7b885a5de797","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"50fb24cd-b977-4df9-a490-f64225fac42e","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"561b816b-fd55-4fc7-b9cd-02dd65433c8d","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"574e4716-037a-4811-a3dd-1a1f206266a9","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"58ab2398-682f-4b01-a663-d719899ca3c9","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"6174aaa8-8897-4e9e-883d-a49fb3bd7c3b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"65c845f0-0cc7-4992-86c3-f4d490f88993","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6c69b838-a08c-4a95-9c34-b92923f2ba4d","difficultyBand":"hard","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"707da3b5-6b4d-43ed-bdd4-8bbd4154d8f9","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"71245a0e-b110-4dbd-8e66-b519f78d5010","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"7b29e5ab-30ed-44ca-8b8f-80a7f8efa445","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"96caaf46-ef14-4263-a873-9218415ea6e8","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"9aadfd8a-f1df-4d17-a73d-c82be4a32f66","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"aa59e030-c4ba-4386-9b05-a4ed52743c47","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"c5160e74-6e5c-4d5f-8399-60a97f2a00d0","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"d3974a89-ab32-442d-ab2e-d3443bcc529d","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"de1331d1-ca2b-48d8-8eac-df6477c5fb69","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"e44f63e4-2574-4b4b-90cd-9c821e6758f1","difficultyBand":"medium","calculatorUse":null,"totalMarks":4,"questionType":"LA","isCalculation":true},{"id":"ea741d59-dafb-496a-af29-6435f92f4b31","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"f2ba430c-e0d4-4a61-b869-ebfba011248a","difficultyBand":"medium","calculatorUse":null,"totalMarks":11,"questionType":"LA","isCalculation":true},{"id":"fb5a3e08-6212-4abb-89c3-747ee401e88b","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true}],"title":"3.2 Arithmetic Series","topicId":"5ef91db9-d11d-45a6-b4f4-105181ee3b51","topicSidebar":"$L3b","topicSlug":"a-level-pure-mathematics-3-2-arithmetic-series","topicTitle":"3.2 Arithmetic Series"}]}],"$L3c","$L3d"]}]

3.2 Arithmetic Series

Question 17

3.2 Arithmetic Series Questions

3.2 Arithmetic Series

Question 17

3.2 Arithmetic Series Questions