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bg-background","children":["$","$L39",null,{"backHref":"/a-level/maths/wjec/questions","basePath":"/a-level/maths/wjec/sequences-and-series/questions","formulaUrl":"https://www.wjec.co.uk/media/4jfdytcz/ss-w27-18-e.pdf","initialIndex":20,"initialQuestionWindow":[{"id":"af8190d3-2895-4d92-a1f1-cd21e0877050","revisionId":"38f5e683-e6ec-4f53-9083-f4cf55e15085","questionNumber":null,"questionType":"LA","specification":"The third term of a geometric series is 18 and the sixth term of the series is 486.","difficulty":2,"difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"50ac4050-ee5a-4e9a-a24f-8d83c2f74dbb","label":"a","content":"Find the common ratio.","markscheme":"- $ar^2 = 18$ and $ar^5 = 486$ M1\n- $r^3 = \\frac{486}{18} = 27 \\implies r = 3$ A1","marks":2,"order":0,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Evidence of set up of two simultaneous equations using the term formula $ar^{n-1}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct value for the common ratio $r$.","code":"A"}],"finalAnswers":["3"],"generatedAt":"2026-05-06T14:05:20.930Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"system_division","label":"Division of Term Equations","steps":[{"type":"text","order":0,"title":"Set up term equations","content":"For a geometric series, the $n$-th term is given by the formula:\n\n$$u_n = a r^{n-1}$$\n\nwhere $a$ is the first term and $r$ is the common ratio.\n\nWe are given:\n- The third term ($u_3$) is $18$\n- The sixth term ($u_6$) is $486$\n\nSubstituting these values into our formula gives us two equations:\n\n$$\\text{Equation 1: } a r^2 = 18$$\n$$\\text{Equation 2: } a r^5 = 486$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{\\htmlData{anchor=eq2}{\\color{@sky}{a r^5}}}{\\htmlData{anchor=eq1}{\\color{@amber}{a r^2}}} = \\frac{\\htmlData{anchor=val2}{\\color{@sky}{486}}}{\\htmlData{anchor=val1}{\\color{@amber}{18}}}$$"},{"latex":"$$$r^3 = 27$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"eq2"},{"color":"amber","style":"text-color","anchor":"eq1"}],"connections":[{"to":"val2","from":"eq2","color":"sky","label":null,"style":"solid"},{"to":"val1","from":"eq1","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Divide the equations","description":"To eliminate the first term $a$, we divide Equation 2 by Equation 1. This cancels out $a$ and simplifies the powers of $r$ using the index law $\\frac{r^5}{r^2} = r^{5-2} = r^3$."},{"type":"text","order":2,"title":"Solve for r","content":"Now, we find the cube root of $27$ to determine the common ratio $r$:\n\n$$r = \\sqrt[3]{27}$$\n\n$$r = 3$$\n\nThus, the common ratio of the geometric series is $3$."}]},{"id":"relative_terms","label":"Relative Term Relationship","steps":[{"type":"text","order":0,"title":"Relate the terms directly","content":"In a geometric sequence, each term is found by multiplying the previous term by a common ratio $r$. This means we can jump directly from the third term, $u_3$, to the sixth term, $u_6$, by multiplying by $r$ three times:\n\n$$u_6 = u_3 \\times r^3$$\n\nThis relative term relationship allows us to find the common ratio $r$ directly without needing to calculate the first term $a$ first.","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"text","order":1,"title":"Substitute the known values","content":"Now we substitute the given values, $u_3 = 18$ and $u_6 = 486$, into our relationship:\n\n$$486 = 18 \\times r^3$$\n\nThis matches the first mark (M1) in the markscheme, which rewards establishing a correct relationship between the terms."},{"type":"text","order":2,"title":"Solve for r","content":"To solve for $r$, we first isolate $r^3$ by dividing both sides by $18$:\n\n$$r^3 = \\frac{486}{18}$$\n\n$$r^3 = 27$$\n\nNow, we take the cube root of both sides to find $r$:\n\n$$r = \\sqrt[3]{27}$$\n\n$$r = 3$$\n\nThis completes the final mark (A1) for finding the correct value of the common ratio.","calculator":[{"gdc":{"expression":"27^(1/3)","instructions":"In the Run-Matrix menu, enter 27^(1/3) or select the cube root function from the math menu, then press [EXE]."},"ti84":{"expression":"³√(27)","instructions":"Press [MATH], select option 4 (³√), enter 27, and press [ENTER]."},"desmos":{"expression":"\\sqrt[3]{27}","instructions":"Type `27^(1/3)` or use the root function with index 3."},"description":"Calculate the cube root of 27 to find the common ratio."}]}]}],"generatedAt":"2026-06-09T08:14:13.219Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"54e532f8-0088-4152-98db-810ab2ebe0a1","label":"b","content":"Find the first term of the series.","markscheme":"- $a(3)^2 = 18 \\implies 9a = 18 \\implies a = 2$ A2","marks":2,"order":1,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Substitution of their value of $r$ into one of the term equations.","code":"M"},{"id":"p2","marks":1,"description":"Correct value for the first term $a$.","code":"A"}],"finalAnswers":["2"],"generatedAt":"2026-05-06T14:05:20.930Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"simultaneous_equations","label":"Simultaneous Equations","steps":[{"type":"text","order":0,"title":"Recall the general formula","content":"To find the terms of a geometric series, we use the general term formula from our studies of geometric sequences:\n\n$$u_n = a r^{n-1}$$\n\nwhere:\n- $a$ is the first term\n- $r$ is the common ratio\n- $n$ is the position of the term"},{"type":"text","order":1,"title":"Set up simultaneous equations","content":"Using the given values from the question, we can write down two equations for $u_3$ and $u_6$:\n\n1. For the third term ($u_3 = 18$):\n $$a r^2 = 18$$\n\n2. For the sixth term ($u_6 = 486$):\n $$a r^5 = 486$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=eq2}{\\color{@amber}{a r^5}}}{\\htmlData{anchor=eq1}{\\color{@sky}{a r^2}}} = \\frac{\\htmlData{anchor=val2}{\\color{@amber}{486}}}{\\htmlData{anchor=val1}{\\color{@sky}{18}}}$"},{"latex":"$$\\htmlData{anchor=result}{r^3} = \t\\htmlData{anchor=val-res}{27}$"},{"latex":"$$r = \\sqrt[3]{27} = 3$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"eq2"},{"color":"sky","style":"text-color","anchor":"eq1"}],"connections":[{"to":"result","from":"eq2","color":"amber","label":"divide terms","style":"solid"},{"to":"val-res","from":"val2","color":"orange","label":"divide values","style":"solid"}]},"order":2,"title":"Solve for common ratio r","description":"We can eliminate the first term $a$ by dividing the second equation by the first equation. This leaves us with a simple single-variable equation for $r$."},{"type":"text","order":3,"title":"Substitute to find term a","content":"Now substitute $r = 3$ back into our first equation to find $a$:\n\n$$\\begin{aligned} a(3)^2 &= 18 \\\\ 9a &= 18 \\\\ a &= 2 \\end{aligned}$$\n\nThus, the first term of the series is **$2$**."},{"type":"text","order":4,"title":"Calculate sum of 10 terms","content":"Now we use the geometric series sum formula:\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nSubstituting our values $a = 2$, $r = 3$, and $n = 10$:\n\n$$\\begin{aligned} S_{10} &= \\frac{2(3^{10} - 1)}{3 - 1} \\\\ S_{10} &= \\frac{2(59049 - 1)}{2} \\\\ S_{10} &= 59048 \\end{aligned}$$\n\nThus, the sum of the first $10$ terms is **$59048$**."}]},{"id":"term_relationships","label":"Term Relationship","steps":[{"type":"text","order":0,"title":"Recall the term relationship","content":"A geometric sequence progresses from one term to the next by multiplying by a constant common ratio, $r$.\n\nInstead of starting from the first term, we can link any two terms in a geometric sequence directly. The sixth term, $u_6$, is $3$ steps away from the third term, $u_3$:\n\n$$u_6 = u_3 \\times r^3$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=u6}{\\color{@purple}{486}} = \\htmlData{anchor=u3}{\\color{@sky}{18}} \\times r^3$"},{"latex":"$$r^3 = \\frac{\\color{@purple}{486}}{\\color{@sky}{18}}$"},{"latex":"$$r^3 = 27 \\implies r = 3$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"u6"},{"color":"sky","style":"text-color","anchor":"u3"}],"connections":[{"to":"u3","from":"u6","color":"indigo","label":"divide by 18","style":"dashed"}]},"order":1,"title":"Calculate the common ratio","description":"Substitute our known terms into the relationship to find $r^3$, then take the cube root to find the common ratio $r$."},{"type":"text","order":2,"title":"Work backward to find a","content":"To find the first term $a$ (which is $u_1$), we can work backward from the third term, $u_3$. Since we multiply by $r$ twice to get from $u_1$ to $u_3$ ($u_3 = a \\times r^2$), we divide $u_3$ by $r^2$ to get back to $a$:\n\n$$a = \\frac{u_3}{r^2}$$\n\nSubstitute $u_3 = 18$ and $r = 3$ into the equation:\n\n$$\\begin{aligned} a &= \\frac{18}{3^2} \\\\ a &= \\frac{18}{9} \\\\ a &= 2 \\end{aligned}$$\n\nThis confirms that the first term of the series is $2$."},{"type":"text","order":3,"title":"Calculate sum of 10 terms","content":"Using the sum formula for a geometric series:\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nSubstitute our values $a = 2$, $r = 3$, and $n = 10$:\n\n$$\\begin{aligned} S_{10} &= \\frac{2(3^{10} - 1)}{3 - 1} \\\\ S_{10} &= 3^{10} - 1 \\\\ S_{10} &= 59048 \\end{aligned}$$\n\nThus, the sum of the first $10$ terms is **$59048$**."}]}],"generatedAt":"2026-06-09T08:15:08.117Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"20bc7002-5f7e-47a8-ab8a-1ba6db594eef","label":"c","content":"Find the sum of the first 10 terms, giving your answer to the nearest whole number.","markscheme":"- $S_{10} = \\frac{2(3^{10}-1)}{3-1}$ M1\n- $S_{10} = 59048$ A1","marks":2,"order":2,"markingRubric":{"totalMarks":2,"points":[{"id":"p1","marks":1,"description":"Correct use of the geometric sum formula with their $a$ and $r$.","code":"M"},{"id":"p2","marks":1,"description":"Final answer to the nearest whole number.","code":"A"}],"finalAnswers":["59048"],"generatedAt":"2026-05-06T14:05:20.930Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"formulaic","label":"Using Sum Formula","steps":[{"type":"text","order":0,"title":"Set up equations","content":"A geometric sequence has terms defined by the formula:\n\n$$u_n = a r^{n-1}$$\n\nwhere $a$ is the first term and $r$ is the common ratio.\n\nUsing the information given in the question, we can write down two simultaneous equations:\n- For the third term:\n $$u_3 = ar^2 = 18$$\n- For the sixth term:\n $$u_6 = ar^5 = 486$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"text","order":1,"title":"Solve for r and a","content":"To eliminate $a$, divide the second equation by the first:\n\n$$\\frac{ar^5}{ar^2} = \\frac{486}{18}$$\n\n$$r^3 = 27$$\n\nTaking the cube root gives:\n\n$$r = 3$$\n\nNow, substitute $r = 3$ back into the equation for the third term to find $a$:\n\n$$a(3)^2 = 18$$\n\n$$9a = 18 \\implies a = 2$$"},{"type":"text","order":2,"title":"Apply sum formula","content":"To find the sum of the first 10 terms ($S_{10}$), we use the geometric series sum formula from your formula booklet:\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nSubstituting our values $a = 2$, $r = 3$, and $n = 10$ into the formula:\n\n$$S_{10} = \\frac{2(3^{10} - 1)}{3 - 1}$$\n\nThis step earns you the first method mark (**M1**).","highlights":[{"text":"Find the sum of the first 10 terms, giving your answer to the nearest whole number.","location":"specification"}]},{"type":"text","order":3,"title":"Calculate the final sum","content":"Simplify the expression to find the final value:\n\n$$S_{10} = \\frac{2(3^{10} - 1)}{2}$$\n\n$$S_{10} = 3^{10} - 1$$\n\n$$S_{10} = 59049 - 1 = 59048$$\n\nThus, the sum of the first 10 terms is **59048** (**A1**)."}]},{"id":"termbyterm","label":"Term-by-Term Summation","steps":[{"type":"text","order":0,"title":"Find the common ratio","content":"From the previous parts of this question, we know the third term ($u_3$) and sixth term ($u_6$) of our geometric sequence are:\n\n$$u_3 = ar^2 = 18$$\n$$u_6 = ar^5 = 486$$\n\nDividing the sixth term by the third term allows us to find the common ratio, $r$:\n\n$$\\frac{ar^5}{ar^2} = \\frac{486}{18}$$\n\n$$r^3 = 27$$\n\n$$r = 3$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"text","order":1,"title":"Find the first term","content":"Now that we know $r = 3$, we can substitute this back into our expression for the third term to find the first term, $a$:\n\n$$a(3)^2 = 18$$\n\n$$9a = 18$$\n\n$$a = 2$$"},{"type":"text","order":2,"title":"List all ten terms","content":"Using our first term $a = 2$ and multiplying by the common ratio $r = 3$ successively, we can generate all $10$ terms of this geometric sequence:\n\n* Term 1: $2$\n* Term 2: $2 \\times 3 = 6$\n* Term 3: $6 \\times 3 = 18$\n* Term 4: $18 \\times 3 = 54$\n* Term 5: $54 \\times 3 = 162$\n* Term 6: $162 \\times 3 = 486$\n* Term 7: $486 \\times 3 = 1458$\n* Term 8: $1458 \\times 3 = 4374$\n* Term 9: $4374 \\times 3 = 13122$\n* Term 10: $13122 \\times 3 = 39366$","highlights":[{"text":"Find the sum of the first 10 terms","location":"specification"}]},{"type":"text","order":3,"title":"Sum the terms directly","content":"To find $S_{10}$, we simply add all ten calculated terms together:\n\n$$S_{10} = 2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 + 13122 + 39366$$\n\n$$S_{10} = 59048$$","calculator":[{"gdc":{"expression":"2+6+18+54+162+486+1458+4374+13122+39366","instructions":"Enter the numbers directly on the main screen and press EXE:\n2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 + 13122 + 39366"},"ti84":{"expression":"2+6+18+54+162+486+1458+4374+13122+39366","instructions":"Enter the numbers directly on the main screen and press ENTER:\n2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 + 13122 + 39366"},"desmos":{"expression":"2+6+18+54+162+486+1458+4374+13122+39366","instructions":"Type the numbers into a new expression line to find the sum."},"description":"Calculate the sum of the sequence directly on your calculator."}]}]}],"generatedAt":"2026-06-09T08:15:46.203Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"50ac4050-ee5a-4e9a-a24f-8d83c2f74dbb:system_division","label":"(a) Division of Term Equations","steps":[{"type":"text","order":0,"title":"Set up term equations","content":"For a geometric series, the $n$-th term is given by the formula:\n\n$$u_n = a r^{n-1}$$\n\nwhere $a$ is the first term and $r$ is the common ratio.\n\nWe are given:\n- The third term ($u_3$) is $18$\n- The sixth term ($u_6$) is $486$\n\nSubstituting these values into our formula gives us two equations:\n\n$$\\text{Equation 1: } a r^2 = 18$$\n$$\\text{Equation 2: } a r^5 = 486$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\frac{\\htmlData{anchor=eq2}{\\color{@sky}{a r^5}}}{\\htmlData{anchor=eq1}{\\color{@amber}{a r^2}}} = \\frac{\\htmlData{anchor=val2}{\\color{@sky}{486}}}{\\htmlData{anchor=val1}{\\color{@amber}{18}}}$$"},{"latex":"$$$r^3 = 27$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"eq2"},{"color":"amber","style":"text-color","anchor":"eq1"}],"connections":[{"to":"val2","from":"eq2","color":"sky","label":null,"style":"solid"},{"to":"val1","from":"eq1","color":"amber","label":null,"style":"solid"}]},"order":1,"title":"Divide the equations","description":"To eliminate the first term $a$, we divide Equation 2 by Equation 1. This cancels out $a$ and simplifies the powers of $r$ using the index law $\\frac{r^5}{r^2} = r^{5-2} = r^3$."},{"type":"text","order":2,"title":"Solve for r","content":"Now, we find the cube root of $27$ to determine the common ratio $r$:\n\n$$r = \\sqrt[3]{27}$$\n\n$$r = 3$$\n\nThus, the common ratio of the geometric series is $3$."}]},{"id":"50ac4050-ee5a-4e9a-a24f-8d83c2f74dbb:relative_terms","label":"(a) Relative Term Relationship","steps":[{"type":"text","order":0,"title":"Relate the terms directly","content":"In a geometric sequence, each term is found by multiplying the previous term by a common ratio $r$. This means we can jump directly from the third term, $u_3$, to the sixth term, $u_6$, by multiplying by $r$ three times:\n\n$$u_6 = u_3 \\times r^3$$\n\nThis relative term relationship allows us to find the common ratio $r$ directly without needing to calculate the first term $a$ first.","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"text","order":1,"title":"Substitute the known values","content":"Now we substitute the given values, $u_3 = 18$ and $u_6 = 486$, into our relationship:\n\n$$486 = 18 \\times r^3$$\n\nThis matches the first mark (M1) in the markscheme, which rewards establishing a correct relationship between the terms."},{"type":"text","order":2,"title":"Solve for r","content":"To solve for $r$, we first isolate $r^3$ by dividing both sides by $18$:\n\n$$r^3 = \\frac{486}{18}$$\n\n$$r^3 = 27$$\n\nNow, we take the cube root of both sides to find $r$:\n\n$$r = \\sqrt[3]{27}$$\n\n$$r = 3$$\n\nThis completes the final mark (A1) for finding the correct value of the common ratio.","calculator":[{"gdc":{"expression":"27^(1/3)","instructions":"In the Run-Matrix menu, enter 27^(1/3) or select the cube root function from the math menu, then press [EXE]."},"ti84":{"expression":"³√(27)","instructions":"Press [MATH], select option 4 (³√), enter 27, and press [ENTER]."},"desmos":{"expression":"\\sqrt[3]{27}","instructions":"Type `27^(1/3)` or use the root function with index 3."},"description":"Calculate the cube root of 27 to find the common ratio."}]}]},{"id":"54e532f8-0088-4152-98db-810ab2ebe0a1:simultaneous_equations","label":"(b) Simultaneous Equations","steps":[{"type":"text","order":0,"title":"Recall the general formula","content":"To find the terms of a geometric series, we use the general term formula from our studies of geometric sequences:\n\n$$u_n = a r^{n-1}$$\n\nwhere:\n- $a$ is the first term\n- $r$ is the common ratio\n- $n$ is the position of the term"},{"type":"text","order":1,"title":"Set up simultaneous equations","content":"Using the given values from the question, we can write down two equations for $u_3$ and $u_6$:\n\n1. For the third term ($u_3 = 18$):\n $$a r^2 = 18$$\n\n2. For the sixth term ($u_6 = 486$):\n $$a r^5 = 486$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=eq2}{\\color{@amber}{a r^5}}}{\\htmlData{anchor=eq1}{\\color{@sky}{a r^2}}} = \\frac{\\htmlData{anchor=val2}{\\color{@amber}{486}}}{\\htmlData{anchor=val1}{\\color{@sky}{18}}}$"},{"latex":"$$\\htmlData{anchor=result}{r^3} = \t\\htmlData{anchor=val-res}{27}$"},{"latex":"$$r = \\sqrt[3]{27} = 3$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"eq2"},{"color":"sky","style":"text-color","anchor":"eq1"}],"connections":[{"to":"result","from":"eq2","color":"amber","label":"divide terms","style":"solid"},{"to":"val-res","from":"val2","color":"orange","label":"divide values","style":"solid"}]},"order":2,"title":"Solve for common ratio r","description":"We can eliminate the first term $a$ by dividing the second equation by the first equation. This leaves us with a simple single-variable equation for $r$."},{"type":"text","order":3,"title":"Substitute to find term a","content":"Now substitute $r = 3$ back into our first equation to find $a$:\n\n$$\\begin{aligned} a(3)^2 &= 18 \\\\ 9a &= 18 \\\\ a &= 2 \\end{aligned}$$\n\nThus, the first term of the series is **$2$**."},{"type":"text","order":4,"title":"Calculate sum of 10 terms","content":"Now we use the geometric series sum formula:\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nSubstituting our values $a = 2$, $r = 3$, and $n = 10$:\n\n$$\\begin{aligned} S_{10} &= \\frac{2(3^{10} - 1)}{3 - 1} \\\\ S_{10} &= \\frac{2(59049 - 1)}{2} \\\\ S_{10} &= 59048 \\end{aligned}$$\n\nThus, the sum of the first $10$ terms is **$59048$**."}]},{"id":"54e532f8-0088-4152-98db-810ab2ebe0a1:term_relationships","label":"(b) Term Relationship","steps":[{"type":"text","order":0,"title":"Recall the term relationship","content":"A geometric sequence progresses from one term to the next by multiplying by a constant common ratio, $r$.\n\nInstead of starting from the first term, we can link any two terms in a geometric sequence directly. The sixth term, $u_6$, is $3$ steps away from the third term, $u_3$:\n\n$$u_6 = u_3 \\times r^3$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=u6}{\\color{@purple}{486}} = \\htmlData{anchor=u3}{\\color{@sky}{18}} \\times r^3$"},{"latex":"$$r^3 = \\frac{\\color{@purple}{486}}{\\color{@sky}{18}}$"},{"latex":"$$r^3 = 27 \\implies r = 3$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"u6"},{"color":"sky","style":"text-color","anchor":"u3"}],"connections":[{"to":"u3","from":"u6","color":"indigo","label":"divide by 18","style":"dashed"}]},"order":1,"title":"Calculate the common ratio","description":"Substitute our known terms into the relationship to find $r^3$, then take the cube root to find the common ratio $r$."},{"type":"text","order":2,"title":"Work backward to find a","content":"To find the first term $a$ (which is $u_1$), we can work backward from the third term, $u_3$. Since we multiply by $r$ twice to get from $u_1$ to $u_3$ ($u_3 = a \\times r^2$), we divide $u_3$ by $r^2$ to get back to $a$:\n\n$$a = \\frac{u_3}{r^2}$$\n\nSubstitute $u_3 = 18$ and $r = 3$ into the equation:\n\n$$\\begin{aligned} a &= \\frac{18}{3^2} \\\\ a &= \\frac{18}{9} \\\\ a &= 2 \\end{aligned}$$\n\nThis confirms that the first term of the series is $2$."},{"type":"text","order":3,"title":"Calculate sum of 10 terms","content":"Using the sum formula for a geometric series:\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nSubstitute our values $a = 2$, $r = 3$, and $n = 10$:\n\n$$\\begin{aligned} S_{10} &= \\frac{2(3^{10} - 1)}{3 - 1} \\\\ S_{10} &= 3^{10} - 1 \\\\ S_{10} &= 59048 \\end{aligned}$$\n\nThus, the sum of the first $10$ terms is **$59048$**."}]},{"id":"20bc7002-5f7e-47a8-ab8a-1ba6db594eef:formulaic","label":"(c) Using Sum Formula","steps":[{"type":"text","order":0,"title":"Set up equations","content":"A geometric sequence has terms defined by the formula:\n\n$$u_n = a r^{n-1}$$\n\nwhere $a$ is the first term and $r$ is the common ratio.\n\nUsing the information given in the question, we can write down two simultaneous equations:\n- For the third term:\n $$u_3 = ar^2 = 18$$\n- For the sixth term:\n $$u_6 = ar^5 = 486$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"text","order":1,"title":"Solve for r and a","content":"To eliminate $a$, divide the second equation by the first:\n\n$$\\frac{ar^5}{ar^2} = \\frac{486}{18}$$\n\n$$r^3 = 27$$\n\nTaking the cube root gives:\n\n$$r = 3$$\n\nNow, substitute $r = 3$ back into the equation for the third term to find $a$:\n\n$$a(3)^2 = 18$$\n\n$$9a = 18 \\implies a = 2$$"},{"type":"text","order":2,"title":"Apply sum formula","content":"To find the sum of the first 10 terms ($S_{10}$), we use the geometric series sum formula from your formula booklet:\n\n$$S_n = \\frac{a(r^n - 1)}{r - 1}$$\n\nSubstituting our values $a = 2$, $r = 3$, and $n = 10$ into the formula:\n\n$$S_{10} = \\frac{2(3^{10} - 1)}{3 - 1}$$\n\nThis step earns you the first method mark (**M1**).","highlights":[{"text":"Find the sum of the first 10 terms, giving your answer to the nearest whole number.","location":"specification"}]},{"type":"text","order":3,"title":"Calculate the final sum","content":"Simplify the expression to find the final value:\n\n$$S_{10} = \\frac{2(3^{10} - 1)}{2}$$\n\n$$S_{10} = 3^{10} - 1$$\n\n$$S_{10} = 59049 - 1 = 59048$$\n\nThus, the sum of the first 10 terms is **59048** (**A1**)."}]},{"id":"20bc7002-5f7e-47a8-ab8a-1ba6db594eef:termbyterm","label":"(c) Term-by-Term Summation","steps":[{"type":"text","order":0,"title":"Find the common ratio","content":"From the previous parts of this question, we know the third term ($u_3$) and sixth term ($u_6$) of our geometric sequence are:\n\n$$u_3 = ar^2 = 18$$\n$$u_6 = ar^5 = 486$$\n\nDividing the sixth term by the third term allows us to find the common ratio, $r$:\n\n$$\\frac{ar^5}{ar^2} = \\frac{486}{18}$$\n\n$$r^3 = 27$$\n\n$$r = 3$$","highlights":[{"text":"third term of a geometric series is 18","location":"specification"},{"text":"sixth term of the series is 486","location":"specification"}]},{"type":"text","order":1,"title":"Find the first term","content":"Now that we know $r = 3$, we can substitute this back into our expression for the third term to find the first term, $a$:\n\n$$a(3)^2 = 18$$\n\n$$9a = 18$$\n\n$$a = 2$$"},{"type":"text","order":2,"title":"List all ten terms","content":"Using our first term $a = 2$ and multiplying by the common ratio $r = 3$ successively, we can generate all $10$ terms of this geometric sequence:\n\n* Term 1: $2$\n* Term 2: $2 \\times 3 = 6$\n* Term 3: $6 \\times 3 = 18$\n* Term 4: $18 \\times 3 = 54$\n* Term 5: $54 \\times 3 = 162$\n* Term 6: $162 \\times 3 = 486$\n* Term 7: $486 \\times 3 = 1458$\n* Term 8: $1458 \\times 3 = 4374$\n* Term 9: $4374 \\times 3 = 13122$\n* Term 10: $13122 \\times 3 = 39366$","highlights":[{"text":"Find the sum of the first 10 terms","location":"specification"}]},{"type":"text","order":3,"title":"Sum the terms directly","content":"To find $S_{10}$, we simply add all ten calculated terms together:\n\n$$S_{10} = 2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 + 13122 + 39366$$\n\n$$S_{10} = 59048$$","calculator":[{"gdc":{"expression":"2+6+18+54+162+486+1458+4374+13122+39366","instructions":"Enter the numbers directly on the main screen and press EXE:\n2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 + 13122 + 39366"},"ti84":{"expression":"2+6+18+54+162+486+1458+4374+13122+39366","instructions":"Enter the numbers directly on the main screen and press ENTER:\n2 + 6 + 18 + 54 + 162 + 486 + 1458 + 4374 + 13122 + 39366"},"desmos":{"expression":"2+6+18+54+162+486+1458+4374+13122+39366","instructions":"Type the numbers into a new expression line to find the sum."},"description":"Calculate the sum of the sequence directly on your calculator."}]}]}],"generatedAt":"2026-06-09T08:15:58.249Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":48},"walkthroughVideo":null},{"id":"b1721fba-4e52-4d3f-ade1-a7f38c98a371","revisionId":"e0ec1ff1-817d-4bc6-a134-723f440894f6","questionNumber":null,"questionType":"LA","specification":"Liam starts a training program where he runs 150 km in the first month and increases his monthly distance by 25 km each subsequent month. Sarah also starts a training program; she runs 210 km in the first month and increases her monthly distance by 15 km each subsequent month.","difficulty":2,"difficultyBand":"medium","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"449b15d7-2935-470a-b355-cfa34ac893fe","label":"a","content":"After how many months have Liam and Sarah run the same total distance.","markscheme":"- Liam: $a=150, d=25$; Sarah: $a=210, d=15$ B1\n- Set sums equal: $\\frac{n}{2}(2(150) + (n-1)25) = \\frac{n}{2}(2(210) + (n-1)15)$ M1\n- $300 + 25n - 25 = 420 + 15n - 15$ M1\n- $10n = 130 \\implies n = 13$ A2","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Identify correct first term (a) and common difference (d) for both sequences.","code":"B"},{"id":"p2","marks":1,"description":"Equate the sum formulas for both arithmetic series.","code":"M"},{"id":"p3","marks":1,"description":"Simplify the sums to a linear equation by canceling n/2 and expanding brackets.","code":"M"},{"id":"p4","marks":2,"description":"Correctly solve the linear equation to find n = 13.","code":"A"}],"finalAnswers":["13"],"generatedAt":"2026-05-06T14:05:13.411Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"algebraic","label":"Algebraic Formula Approach","steps":[{"type":"text","order":0,"title":"Identify arithmetic series parameters","content":"To find when Liam and Sarah have run the same total distance, we first recognize that both of their monthly training programs form an **arithmetic progression (AP)**.\n\nWe need to find the first term ($a$) and the common difference ($d$) for both runners:\n- **Liam** starts with $150\\text{ km}$ and increases by $25\\text{ km}$ each month:\n $$a_L = 150, \\quad d_L = 25$$\n- **Sarah** starts with $210\\text{ km}$ and increases by $15\\text{ km}$ each month:\n $$a_S = 210, \\quad d_S = 15$$","highlights":[{"text":"Liam starts a training program where he runs 150 km in the first month and increases his monthly distance by 25 km each subsequent month.","location":"specification"},{"text":"Sarah also starts a training program; she runs 210 km in the first month and increases her monthly distance by 15 km each subsequent month.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"The formula for the sum of the first $n$ terms of an arithmetic progression is:"},{"latex":"$$$S_n = \\frac{n}{2}\\left(2a + (n-1)d\\right)$$"},{"text":"We equate the total distance run by Liam ($S_{n,\\text{Liam}}$) and Sarah ($S_{n,\\text{Sarah}}$) over $n$ months:"},{"latex":"$$$\\frac{n}{2}\\left(2(\\htmlData{anchor=a-liam}{\\color{@sky}{150}}) + (n-1)\\htmlData{anchor=d-liam}{\\color{@amber}{25}}\\right) = \\frac{n}{2}\\left(2(\\htmlData{anchor=a-sarah}{\\color{@pink}{210}}) + (n-1)\\htmlData{anchor=d-sarah}{\\color{@purple}{15}}\\right)$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-liam"},{"color":"amber","style":"text-color","anchor":"d-liam"},{"color":"pink","style":"text-color","anchor":"a-sarah"},{"color":"purple","style":"text-color","anchor":"d-sarah"}],"connections":[]},"order":1,"title":"Set up sum equations","description":"We substitute the values of $a$ and $d$ for both runners into the arithmetic series sum formula."},{"type":"text","order":2,"title":"Simplify the equation","content":"Since $n \\neq 0$ (as they run for at least one month), we can simplify the equation by dividing both sides by $\\frac{n}{2}$:\n\n$$2(150) + 25(n-1) = 2(210) + 15(n-1)$$\n\nNow, expand the brackets:\n\n$$300 + 25n - 25 = 420 + 15n - 15$$"},{"type":"text","order":3,"title":"Solve for n","content":"Group the terms containing $n$ on one side and the constant numbers on the other:\n\n$$\\begin{aligned} 275 + 25n &= 405 + 15n \\\\ 25n - 15n &= 405 - 275 \\\\ 10n &= 130 \\\\ n &= 13 \\end{aligned}$$\n\nThus, after **13 months**, Liam and Sarah have run the same total distance."}]},{"id":"gdc_intersection","label":"Graphing Calculator Approach","steps":[{"type":"text","order":0,"title":"Analyze the arithmetic series","content":"To find when Liam and Sarah have run the same total distance, we need to represent the cumulative sum of distance run by each person as an arithmetic series.\n\nFor an arithmetic series, the sum of the first $n$ terms is given by the formula:\n$$S_n = \\frac{n}{2}(2a + (n-1)d)$$\n\nwhere $a$ is the first term (distance in month 1) and $d$ is the common difference (monthly increase).\n\nLet's write down the parameters for both Liam and Sarah:\n- **Liam**: first term $a_L = 150$, difference $d_L = 25$\n- **Sarah**: first term $a_S = 210$, difference $d_S = 15$","highlights":[{"text":"runs 150 km in the first month and increases his monthly distance by 25 km","location":"specification"},{"text":"runs 210 km in the first month and increases her monthly distance by 15 km","location":"specification"}]},{"type":"text","order":1,"title":"Set up the functions","content":"We can set up two functions representing the cumulative total run by each person after $x$ months. Let $x$ represent the number of months:\n\n- **Liam's total distance function** ($Y_1$):\n$$Y_1 = \\frac{x}{2}(2(150) + (x-1)25) = \\frac{x}{2}(300 + 25x - 25) = \\frac{x}{2}(275 + 25x)$$\n\n- **Sarah's total distance function** ($Y_2$):\n$$Y_2 = \\frac{x}{2}(2(210) + (x-1)15) = \\frac{x}{2}(420 + 15x - 15) = \\frac{x}{2}(405 + 15x)$$"},{"type":"text","order":2,"title":"Use a graphing calculator","content":"To find the number of months when their total distances are the same, we want to find the point of intersection of $Y_1$ and $Y_2$ for $x > 0$. We can enter these equations into a graphing calculator and find where they intersect, or check the table of values.","calculator":[{"gdc":{"expression":"Y1 = (x/2)*(300 + 25*(x-1))\nY2 = (x/2)*(420 + 15*(x-1))","instructions":"1. Go to the Graph menu and define:\n Y1 = (x / 2) * (300 + 25 * (x - 1))\n Y2 = (x / 2) * (420 + 15 * (x - 1))\n2. View the Table of values or use the G-Solve feature.\n3. In G-Solve, select 'Intersection' (INTSECT) to find the coordinate where the curves meet."},"ti84":{"expression":"Y1 = (X/2)*(300 + 25*(X-1))\nY2 = (X/2)*(420 + 15*(X-1))","instructions":"1. Press [Y=] and enter:\n Y1 = (X / 2) * (2 * 150 + (X - 1) * 25)\n Y2 = (X / 2) * (2 * 210 + (X - 1) * 15)\n2. Press [2nd] [TABLE] (above [GRAPH]).\n3. Scroll down to find the row where Y1 and Y2 have the same value.\n Alternatively, press [WINDOW] and set:\n Xmin = 0, Xmax = 20, Ymin = 0, Ymax = 5000.\n4. Press [GRAPH], then [2nd] [CALC] and select '5: intersect'. Select the first curve, second curve, and press [ENTER] to guess."},"desmos":{"expression":"y = (x/2)*(300 + 25*(x-1))\ny = (x/2)*(420 + 15*(x-1))","instructions":"1. Enter the two equations in the input lines:\n y = (x / 2) * (2 * 150 + (x - 1) * 25)\n y = (x / 2) * (2 * 210 + (x - 1) * 15)\n2. Zoom out or adjust the window axes to see the intersection point in the first quadrant.\n3. Click on the point of intersection to reveal its coordinates."},"description":"Enter the functions and find their intersection or check the table of values."}]},{"type":"text","order":3,"title":"Interpret the results","content":"Looking at our calculator's table or finding the intersection point, we get:\n- At $x = 13$, the values of both functions are:\n$$Y_1 = \\frac{13}{2}(300 + 25(12)) = 6.5 \\times 600 = 3900 \\text{ km}$$\n$$Y_2 = \\frac{13}{2}(420 + 15(12)) = 6.5 \\times 600 = 3900 \\text{ km}$$\n\nThus, the intersection point is $(13, 3900)$. This means that after $13$ months, both Liam and Sarah have run the same total distance of $3900$ km."}]},{"id":"table_listing","label":"Iterative Calculation Approach","steps":[{"type":"text","order":0,"title":"Understand the approach","content":"To find out when Liam and Sarah have run the same total distance, we will list their monthly distances and cumulative (running) total distances month-by-month. We will continue this systematic calculation until their cumulative totals match.","highlights":[{"text":"runs 150 km in the first month and increases his monthly distance by 25 km each subsequent month","location":"specification"},{"text":"runs 210 km in the first month and increases her monthly distance by 15 km each subsequent month","location":"specification"}]},{"type":"text","order":1,"title":"Calculate Months 1 to 4","content":"Let's calculate the distances run each month and the running totals for Liam and Sarah starting from Month 1:\n\n* **Liam**: Starts at $150\\text{ km}$ and adds $25\\text{ km}$ each month.\n* **Sarah**: Starts at $210\\text{ km}$ and adds $15\\text{ km}$ each month.\n\n| Month | Liam's Distance (km) | Liam's Total (km) | Sarah's Distance (km) | Sarah's Total (km) |\n| :---: | :---: | :---: | :---: | :---: |\n| 1 | 150 | 150 | 210 | 210 |\n| 2 | 175 | 325 | 225 | 435 |\n| 3 | 200 | 525 | 240 | 675 |\n| 4 | 225 | 750 | 255 | 930 |\n\nCurrently, Sarah is in the lead with a total of $930\\text{ km}$ compared to Liam's $750\\text{ km}$."},{"type":"text","order":2,"title":"Calculate Months 5 to 8","content":"Let's continue adding the monthly increments and accumulating the totals for both runners:\n\n| Month | Liam's Distance (km) | Liam's Total (km) | Sarah's Distance (km) | Sarah's Total (km) |\n| :---: | :---: | :---: | :---: | :---: |\n| 5 | 250 | 1000 | 270 | 1200 |\n| 6 | 275 | 1275 | 285 | 1485 |\n| 7 | 300 | 1575 | 300 | 1785 |\n| 8 | 325 | 1900 | 315 | 2100 |\n\nNotice that at Month 7, both run the same monthly distance ($300\\text{ km}$), but their cumulative totals are not yet equal ($1575\\text{ km}$ vs $1785\\text{ km}$). From Month 8 onward, Liam starts running more per month than Sarah, allowing him to catch up."},{"type":"text","order":3,"title":"Calculate Months 9 to 13","content":"Continuing the calculations until the cumulative totals match:\n\n| Month | Liam's Distance (km) | Liam's Total (km) | Sarah's Distance (km) | Sarah's Total (km) |\n| :---: | :---: | :---: | :---: | :---: |\n| 9 | 350 | 2250 | 330 | 2430 |\n| 10 | 375 | 2625 | 345 | 2775 |\n| 11 | 400 | 3025 | 360 | 3135 |\n| 12 | 425 | 3450 | 375 | 3510 |\n| 13 | 450 | **3900** | 390 | **3900** |\n\nAt **Month 13**, both Liam and Sarah have run the exact same cumulative total distance of $3900\\text{ km}$."},{"type":"text","order":4,"title":"State the final answer","content":"Therefore, Liam and Sarah have run the same total distance after **13** months."}]}],"generatedAt":"2026-06-09T09:04:31.602Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"449b15d7-2935-470a-b355-cfa34ac893fe:algebraic","label":"Algebraic Formula Approach","steps":[{"type":"text","order":0,"title":"Identify arithmetic series parameters","content":"To find when Liam and Sarah have run the same total distance, we first recognize that both of their monthly training programs form an **arithmetic progression (AP)**.\n\nWe need to find the first term ($a$) and the common difference ($d$) for both runners:\n- **Liam** starts with $150\\text{ km}$ and increases by $25\\text{ km}$ each month:\n $$a_L = 150, \\quad d_L = 25$$\n- **Sarah** starts with $210\\text{ km}$ and increases by $15\\text{ km}$ each month:\n $$a_S = 210, \\quad d_S = 15$$","highlights":[{"text":"Liam starts a training program where he runs 150 km in the first month and increases his monthly distance by 25 km each subsequent month.","location":"specification"},{"text":"Sarah also starts a training program; she runs 210 km in the first month and increases her monthly distance by 15 km each subsequent month.","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"text":"The formula for the sum of the first $n$ terms of an arithmetic progression is:"},{"latex":"$$$S_n = \\frac{n}{2}\\left(2a + (n-1)d\\right)$$"},{"text":"We equate the total distance run by Liam ($S_{n,\\text{Liam}}$) and Sarah ($S_{n,\\text{Sarah}}$) over $n$ months:"},{"latex":"$$$\\frac{n}{2}\\left(2(\\htmlData{anchor=a-liam}{\\color{@sky}{150}}) + (n-1)\\htmlData{anchor=d-liam}{\\color{@amber}{25}}\\right) = \\frac{n}{2}\\left(2(\\htmlData{anchor=a-sarah}{\\color{@pink}{210}}) + (n-1)\\htmlData{anchor=d-sarah}{\\color{@purple}{15}}\\right)$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-liam"},{"color":"amber","style":"text-color","anchor":"d-liam"},{"color":"pink","style":"text-color","anchor":"a-sarah"},{"color":"purple","style":"text-color","anchor":"d-sarah"}],"connections":[]},"order":1,"title":"Set up sum equations","description":"We substitute the values of $a$ and $d$ for both runners into the arithmetic series sum formula."},{"type":"text","order":2,"title":"Simplify the equation","content":"Since $n \\neq 0$ (as they run for at least one month), we can simplify the equation by dividing both sides by $\\frac{n}{2}$:\n\n$$2(150) + 25(n-1) = 2(210) + 15(n-1)$$\n\nNow, expand the brackets:\n\n$$300 + 25n - 25 = 420 + 15n - 15$$"},{"type":"text","order":3,"title":"Solve for n","content":"Group the terms containing $n$ on one side and the constant numbers on the other:\n\n$$\\begin{aligned} 275 + 25n &= 405 + 15n \\\\ 25n - 15n &= 405 - 275 \\\\ 10n &= 130 \\\\ n &= 13 \\end{aligned}$$\n\nThus, after **13 months**, Liam and Sarah have run the same total distance."}]},{"id":"449b15d7-2935-470a-b355-cfa34ac893fe:gdc_intersection","label":"Graphing Calculator Approach","steps":[{"type":"text","order":0,"title":"Analyze the arithmetic series","content":"To find when Liam and Sarah have run the same total distance, we need to represent the cumulative sum of distance run by each person as an arithmetic series.\n\nFor an arithmetic series, the sum of the first $n$ terms is given by the formula:\n$$S_n = \\frac{n}{2}(2a + (n-1)d)$$\n\nwhere $a$ is the first term (distance in month 1) and $d$ is the common difference (monthly increase).\n\nLet's write down the parameters for both Liam and Sarah:\n- **Liam**: first term $a_L = 150$, difference $d_L = 25$\n- **Sarah**: first term $a_S = 210$, difference $d_S = 15$","highlights":[{"text":"runs 150 km in the first month and increases his monthly distance by 25 km","location":"specification"},{"text":"runs 210 km in the first month and increases her monthly distance by 15 km","location":"specification"}]},{"type":"text","order":1,"title":"Set up the functions","content":"We can set up two functions representing the cumulative total run by each person after $x$ months. Let $x$ represent the number of months:\n\n- **Liam's total distance function** ($Y_1$):\n$$Y_1 = \\frac{x}{2}(2(150) + (x-1)25) = \\frac{x}{2}(300 + 25x - 25) = \\frac{x}{2}(275 + 25x)$$\n\n- **Sarah's total distance function** ($Y_2$):\n$$Y_2 = \\frac{x}{2}(2(210) + (x-1)15) = \\frac{x}{2}(420 + 15x - 15) = \\frac{x}{2}(405 + 15x)$$"},{"type":"text","order":2,"title":"Use a graphing calculator","content":"To find the number of months when their total distances are the same, we want to find the point of intersection of $Y_1$ and $Y_2$ for $x > 0$. We can enter these equations into a graphing calculator and find where they intersect, or check the table of values.","calculator":[{"gdc":{"expression":"Y1 = (x/2)*(300 + 25*(x-1))\nY2 = (x/2)*(420 + 15*(x-1))","instructions":"1. Go to the Graph menu and define:\n Y1 = (x / 2) * (300 + 25 * (x - 1))\n Y2 = (x / 2) * (420 + 15 * (x - 1))\n2. View the Table of values or use the G-Solve feature.\n3. In G-Solve, select 'Intersection' (INTSECT) to find the coordinate where the curves meet."},"ti84":{"expression":"Y1 = (X/2)*(300 + 25*(X-1))\nY2 = (X/2)*(420 + 15*(X-1))","instructions":"1. Press [Y=] and enter:\n Y1 = (X / 2) * (2 * 150 + (X - 1) * 25)\n Y2 = (X / 2) * (2 * 210 + (X - 1) * 15)\n2. Press [2nd] [TABLE] (above [GRAPH]).\n3. Scroll down to find the row where Y1 and Y2 have the same value.\n Alternatively, press [WINDOW] and set:\n Xmin = 0, Xmax = 20, Ymin = 0, Ymax = 5000.\n4. Press [GRAPH], then [2nd] [CALC] and select '5: intersect'. Select the first curve, second curve, and press [ENTER] to guess."},"desmos":{"expression":"y = (x/2)*(300 + 25*(x-1))\ny = (x/2)*(420 + 15*(x-1))","instructions":"1. Enter the two equations in the input lines:\n y = (x / 2) * (2 * 150 + (x - 1) * 25)\n y = (x / 2) * (2 * 210 + (x - 1) * 15)\n2. Zoom out or adjust the window axes to see the intersection point in the first quadrant.\n3. Click on the point of intersection to reveal its coordinates."},"description":"Enter the functions and find their intersection or check the table of values."}]},{"type":"text","order":3,"title":"Interpret the results","content":"Looking at our calculator's table or finding the intersection point, we get:\n- At $x = 13$, the values of both functions are:\n$$Y_1 = \\frac{13}{2}(300 + 25(12)) = 6.5 \\times 600 = 3900 \\text{ km}$$\n$$Y_2 = \\frac{13}{2}(420 + 15(12)) = 6.5 \\times 600 = 3900 \\text{ km}$$\n\nThus, the intersection point is $(13, 3900)$. This means that after $13$ months, both Liam and Sarah have run the same total distance of $3900$ km."}]},{"id":"449b15d7-2935-470a-b355-cfa34ac893fe:table_listing","label":"Iterative Calculation Approach","steps":[{"type":"text","order":0,"title":"Understand the approach","content":"To find out when Liam and Sarah have run the same total distance, we will list their monthly distances and cumulative (running) total distances month-by-month. We will continue this systematic calculation until their cumulative totals match.","highlights":[{"text":"runs 150 km in the first month and increases his monthly distance by 25 km each subsequent month","location":"specification"},{"text":"runs 210 km in the first month and increases her monthly distance by 15 km each subsequent month","location":"specification"}]},{"type":"text","order":1,"title":"Calculate Months 1 to 4","content":"Let's calculate the distances run each month and the running totals for Liam and Sarah starting from Month 1:\n\n* **Liam**: Starts at $150\\text{ km}$ and adds $25\\text{ km}$ each month.\n* **Sarah**: Starts at $210\\text{ km}$ and adds $15\\text{ km}$ each month.\n\n| Month | Liam's Distance (km) | Liam's Total (km) | Sarah's Distance (km) | Sarah's Total (km) |\n| :---: | :---: | :---: | :---: | :---: |\n| 1 | 150 | 150 | 210 | 210 |\n| 2 | 175 | 325 | 225 | 435 |\n| 3 | 200 | 525 | 240 | 675 |\n| 4 | 225 | 750 | 255 | 930 |\n\nCurrently, Sarah is in the lead with a total of $930\\text{ km}$ compared to Liam's $750\\text{ km}$."},{"type":"text","order":2,"title":"Calculate Months 5 to 8","content":"Let's continue adding the monthly increments and accumulating the totals for both runners:\n\n| Month | Liam's Distance (km) | Liam's Total (km) | Sarah's Distance (km) | Sarah's Total (km) |\n| :---: | :---: | :---: | :---: | :---: |\n| 5 | 250 | 1000 | 270 | 1200 |\n| 6 | 275 | 1275 | 285 | 1485 |\n| 7 | 300 | 1575 | 300 | 1785 |\n| 8 | 325 | 1900 | 315 | 2100 |\n\nNotice that at Month 7, both run the same monthly distance ($300\\text{ km}$), but their cumulative totals are not yet equal ($1575\\text{ km}$ vs $1785\\text{ km}$). From Month 8 onward, Liam starts running more per month than Sarah, allowing him to catch up."},{"type":"text","order":3,"title":"Calculate Months 9 to 13","content":"Continuing the calculations until the cumulative totals match:\n\n| Month | Liam's Distance (km) | Liam's Total (km) | Sarah's Distance (km) | Sarah's Total (km) |\n| :---: | :---: | :---: | :---: | :---: |\n| 9 | 350 | 2250 | 330 | 2430 |\n| 10 | 375 | 2625 | 345 | 2775 |\n| 11 | 400 | 3025 | 360 | 3135 |\n| 12 | 425 | 3450 | 375 | 3510 |\n| 13 | 450 | **3900** | 390 | **3900** |\n\nAt **Month 13**, both Liam and Sarah have run the exact same cumulative total distance of $3900\\text{ km}$."},{"type":"text","order":4,"title":"State the final answer","content":"Therefore, Liam and Sarah have run the same total distance after **13** months."}]}],"generatedAt":"2026-06-09T09:05:17.551Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":36},"walkthroughVideo":null},{"id":"c1feb070-8b01-4aab-a059-0b23dc16bc0f","revisionId":"1ec54fc5-3381-4c33-8165-e7f200955a2a","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"medium","tier":null,"totalMarks":3,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"344a2b03-b6e3-473b-aa41-f7345f21cc22","label":"a","content":"Work out the first four terms, in ascending powers of $x$, in the binomial expansion of $\\frac{1}{(1+x)^3}$","markscheme":"- Uses the formula $(1+x)^n = 1 + nx + \\frac{n(n-1)}{2!}x^2 + \\frac{n(n-1)(n-2)}{3!}x^3$ with $n = -3$ M1\n- Correct substitution: $1 + (-3)x + \\frac{(-3)(-4)}{2}x^2 + \\frac{(-3)(-4)(-5)}{6}x^3$ M1\n- Simplifies to $1 - 3x + 6x^2 - 10x^3$ A1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Uses the binomial expansion formula for a negative power with $n = -3$","code":"M"},{"id":"p2","marks":1,"description":"Correct substitution of $n = -3$ into the binomial formula for at least the first three terms.","code":"M"},{"id":"p3","marks":1,"description":"Simplifies correctly to obtain $1 - 3x + 6x^2 - 10x^3$.","code":"A"}],"finalAnswers":["1 - 3x + 6x^2 - 10x^3"],"generatedAt":"2026-05-06T14:03:53.797Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"binomial_formula","label":"Binomial Series Formula","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"To expand $\\frac{1}{(1+x)^3}$ using the binomial series formula, we first rewrite the fraction with a negative index:\n\n$$\\frac{1}{(1+x)^3} = (1+x)^{-3}$$\n\nThis puts our expression into the standard form $(1+x)^n$ where $n = -3$."},{"type":"hyper_latex","block":{"nodes":[{"text":"Recall the general binomial expansion formula for any rational power $n$:"},{"latex":"$$$(1+x)^{\\htmlData{anchor=n-val}{\\color{@sky}{n}}} = 1 + \\htmlData{anchor=n1}{\\color{@sky}{n}}x + \\frac{\\htmlData{anchor=n2}{\\color{@sky}{n}}(\\htmlData{anchor=n3}{\\color{@sky}{n}}-1)}{2!}x^2 + \\frac{\\htmlData{anchor=n4}{\\color{@sky}{n}}(\\htmlData{anchor=n5}{\\color{@sky}{n}}-1)(\\htmlData{anchor=n6}{\\color{@sky}{n}}-2)}{3!}x^3 + \\dots$$"},{"text":"For our problem, we substitute the value of $n$:"},{"latex":"$$$\\htmlData{anchor=sub-val}{\\color{@sky}{n = -3}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"sky","style":"text-color","anchor":"sub-val"}],"connections":[{"to":"sub-val","from":"n-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Apply binomial series formula","description":"We substitute $n = -3$ into the general binomial expansion formula."},{"type":"text","order":2,"title":"Substitute and simplify","content":"Substituting $n = -3$ into the expansion formula gives:\n\n$$(1+x)^{-3} = 1 + (-3)x + \\frac{(-3)(-4)}{2!}x^2 + \\frac{(-3)(-4)(-5)}{3!}x^3 + \\dots$$\n\nNow, let's calculate the coefficient of each term step-by-step:\n\n- **$x$ term:** $(-3)x = -3x$\n- **$x^2$ term:** $\\frac{(-3)(-4)}{2}x^2 = \\frac{12}{2}x^2 = 6x^2$\n- **$x^3$ term:** $\\frac{(-3)(-4)(-5)}{6}x^3 = \\frac{-60}{6}x^3 = -10x^3$\n\nPutting it all together, we get:\n\n$$(1+x)^{-3} \\approx 1 - 3x + 6x^2 - 10x^3$$"}]},{"id":"maclaurin_series","label":"Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"Define the function","content":"To find the expansion using the Maclaurin series method, we first rewrite the given expression as a function $f(x)$ with a negative exponent:\n\n$$f(x) = \\frac{1}{(1+x)^3} = (1+x)^{-3}$$"},{"type":"text","order":1,"title":"Find the derivatives","content":"Next, we find the first three derivatives of $f(x)$ using the chain rule and power rule:\n\n$$\\begin{aligned} f'(x) &= -3(1+x)^{-4} \\\\ f''(x) &= (-3)(-4)(1+x)^{-5} = 12(1+x)^{-5} \\\\ f'''(x) &= 12(-5)(1+x)^{-6} = -60(1+x)^{-6} \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(\\htmlData{anchor=x0}{\\color{@sky}{0}}) = (1+\\color{@sky}{0})^{-3} = \\htmlData{anchor=f0}{\\color{@pink}{1}}$"},{"latex":"$$f'(\\color{@sky}{0}) = -3(1+\\color{@sky}{0})^{-4} = \\htmlData{anchor=f1}{\\color{@purple}{-3}}$"},{"latex":"$$f''(\\color{@sky}{0}) = 12(1+\\color{@sky}{0})^{-5} = \\htmlData{anchor=f2}{\\color{@green}{12}}$"},{"latex":"$$f'''(\\color{@sky}{0}) = -60(1+\\color{@sky}{0})^{-6} = \\htmlData{anchor=f3}{\\color{@amber}{-60}}$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"f0"},{"color":"purple","style":"text-color","anchor":"f1"},{"color":"green","style":"text-color","anchor":"f2"},{"color":"amber","style":"text-color","anchor":"f3"}],"connections":[]},"order":2,"title":"Evaluate at $x = 0$","description":"Now, we evaluate the function and each of its derivatives at $x = 0$ to find the coefficients of our series."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) \\approx \\htmlData{anchor=f0-term}{\\color{@pink}{f(0)}} + \\htmlData{anchor=f1-term}{\\color{@purple}{f'(0)}}x + \\frac{\\htmlData{anchor=f2-term}{\\color{@green}{f''(0)}}}{2!}x^2 + \\frac{\\htmlData{anchor=f3-term}{\\color{@amber}{f'''(0)}}}{3!}x^3$"},{"latex":"$$f(x) \\approx \\htmlData{anchor=f0-sub}{\\color{@pink}{1}} + (\\htmlData{anchor=f1-sub}{\\color{@purple}{-3}})x + \\frac{\\htmlData{anchor=f2-sub}{\\color{@green}{12}}}{2}x^2 + \\frac{\\htmlData{anchor=f3-sub}{\\color{@amber}{-60}}}{6}x^3$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"f0-term"},{"color":"purple","style":"text-color","anchor":"f1-term"},{"color":"green","style":"text-color","anchor":"f2-term"},{"color":"amber","style":"text-color","anchor":"f3-term"},{"color":"pink","style":"text-color","anchor":"f0-sub"},{"color":"purple","style":"text-color","anchor":"f1-sub"},{"color":"green","style":"text-color","anchor":"f2-sub"},{"color":"amber","style":"text-color","anchor":"f3-sub"}],"connections":[{"to":"f0-sub","from":"f0-term","color":"pink","label":null,"style":"solid"},{"to":"f1-sub","from":"f1-term","color":"purple","label":null,"style":"solid"},{"to":"f2-sub","from":"f2-term","color":"green","label":null,"style":"solid"},{"to":"f3-sub","from":"f3-term","color":"amber","label":null,"style":"solid"}]},"order":3,"title":"Apply Maclaurin formula","description":"The Maclaurin series expansion formula for the first four terms is given by:"},{"type":"text","order":4,"title":"Simplify final terms","content":"Simplifying each coefficient gives us our final ascending series expansion:\n\n$$f(x) \\approx 1 - 3x + 6x^2 - 10x^3$$"}]}],"generatedAt":"2026-06-09T09:45:29.761Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"344a2b03-b6e3-473b-aa41-f7345f21cc22:binomial_formula","label":"Binomial Series Formula","steps":[{"type":"text","order":0,"title":"Rewrite the expression","content":"To expand $\\frac{1}{(1+x)^3}$ using the binomial series formula, we first rewrite the fraction with a negative index:\n\n$$\\frac{1}{(1+x)^3} = (1+x)^{-3}$$\n\nThis puts our expression into the standard form $(1+x)^n$ where $n = -3$."},{"type":"hyper_latex","block":{"nodes":[{"text":"Recall the general binomial expansion formula for any rational power $n$:"},{"latex":"$$$(1+x)^{\\htmlData{anchor=n-val}{\\color{@sky}{n}}} = 1 + \\htmlData{anchor=n1}{\\color{@sky}{n}}x + \\frac{\\htmlData{anchor=n2}{\\color{@sky}{n}}(\\htmlData{anchor=n3}{\\color{@sky}{n}}-1)}{2!}x^2 + \\frac{\\htmlData{anchor=n4}{\\color{@sky}{n}}(\\htmlData{anchor=n5}{\\color{@sky}{n}}-1)(\\htmlData{anchor=n6}{\\color{@sky}{n}}-2)}{3!}x^3 + \\dots$$"},{"text":"For our problem, we substitute the value of $n$:"},{"latex":"$$$\\htmlData{anchor=sub-val}{\\color{@sky}{n = -3}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"n-val"},{"color":"sky","style":"text-color","anchor":"sub-val"}],"connections":[{"to":"sub-val","from":"n-val","color":"sky","label":"substitute","style":"solid"}]},"order":1,"title":"Apply binomial series formula","description":"We substitute $n = -3$ into the general binomial expansion formula."},{"type":"text","order":2,"title":"Substitute and simplify","content":"Substituting $n = -3$ into the expansion formula gives:\n\n$$(1+x)^{-3} = 1 + (-3)x + \\frac{(-3)(-4)}{2!}x^2 + \\frac{(-3)(-4)(-5)}{3!}x^3 + \\dots$$\n\nNow, let's calculate the coefficient of each term step-by-step:\n\n- **$x$ term:** $(-3)x = -3x$\n- **$x^2$ term:** $\\frac{(-3)(-4)}{2}x^2 = \\frac{12}{2}x^2 = 6x^2$\n- **$x^3$ term:** $\\frac{(-3)(-4)(-5)}{6}x^3 = \\frac{-60}{6}x^3 = -10x^3$\n\nPutting it all together, we get:\n\n$$(1+x)^{-3} \\approx 1 - 3x + 6x^2 - 10x^3$$"}]},{"id":"344a2b03-b6e3-473b-aa41-f7345f21cc22:maclaurin_series","label":"Maclaurin Series Expansion","steps":[{"type":"text","order":0,"title":"Define the function","content":"To find the expansion using the Maclaurin series method, we first rewrite the given expression as a function $f(x)$ with a negative exponent:\n\n$$f(x) = \\frac{1}{(1+x)^3} = (1+x)^{-3}$$"},{"type":"text","order":1,"title":"Find the derivatives","content":"Next, we find the first three derivatives of $f(x)$ using the chain rule and power rule:\n\n$$\\begin{aligned} f'(x) &= -3(1+x)^{-4} \\\\ f''(x) &= (-3)(-4)(1+x)^{-5} = 12(1+x)^{-5} \\\\ f'''(x) &= 12(-5)(1+x)^{-6} = -60(1+x)^{-6} \\end{aligned}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(\\htmlData{anchor=x0}{\\color{@sky}{0}}) = (1+\\color{@sky}{0})^{-3} = \\htmlData{anchor=f0}{\\color{@pink}{1}}$"},{"latex":"$$f'(\\color{@sky}{0}) = -3(1+\\color{@sky}{0})^{-4} = \\htmlData{anchor=f1}{\\color{@purple}{-3}}$"},{"latex":"$$f''(\\color{@sky}{0}) = 12(1+\\color{@sky}{0})^{-5} = \\htmlData{anchor=f2}{\\color{@green}{12}}$"},{"latex":"$$f'''(\\color{@sky}{0}) = -60(1+\\color{@sky}{0})^{-6} = \\htmlData{anchor=f3}{\\color{@amber}{-60}}$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"f0"},{"color":"purple","style":"text-color","anchor":"f1"},{"color":"green","style":"text-color","anchor":"f2"},{"color":"amber","style":"text-color","anchor":"f3"}],"connections":[]},"order":2,"title":"Evaluate at $x = 0$","description":"Now, we evaluate the function and each of its derivatives at $x = 0$ to find the coefficients of our series."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$f(x) \\approx \\htmlData{anchor=f0-term}{\\color{@pink}{f(0)}} + \\htmlData{anchor=f1-term}{\\color{@purple}{f'(0)}}x + \\frac{\\htmlData{anchor=f2-term}{\\color{@green}{f''(0)}}}{2!}x^2 + \\frac{\\htmlData{anchor=f3-term}{\\color{@amber}{f'''(0)}}}{3!}x^3$"},{"latex":"$$f(x) \\approx \\htmlData{anchor=f0-sub}{\\color{@pink}{1}} + (\\htmlData{anchor=f1-sub}{\\color{@purple}{-3}})x + \\frac{\\htmlData{anchor=f2-sub}{\\color{@green}{12}}}{2}x^2 + \\frac{\\htmlData{anchor=f3-sub}{\\color{@amber}{-60}}}{6}x^3$"}],"highlights":[{"color":"pink","style":"text-color","anchor":"f0-term"},{"color":"purple","style":"text-color","anchor":"f1-term"},{"color":"green","style":"text-color","anchor":"f2-term"},{"color":"amber","style":"text-color","anchor":"f3-term"},{"color":"pink","style":"text-color","anchor":"f0-sub"},{"color":"purple","style":"text-color","anchor":"f1-sub"},{"color":"green","style":"text-color","anchor":"f2-sub"},{"color":"amber","style":"text-color","anchor":"f3-sub"}],"connections":[{"to":"f0-sub","from":"f0-term","color":"pink","label":null,"style":"solid"},{"to":"f1-sub","from":"f1-term","color":"purple","label":null,"style":"solid"},{"to":"f2-sub","from":"f2-term","color":"green","label":null,"style":"solid"},{"to":"f3-sub","from":"f3-term","color":"amber","label":null,"style":"solid"}]},"order":3,"title":"Apply Maclaurin formula","description":"The Maclaurin series expansion formula for the first four terms is 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Sequences and Series (A-level only)","topicId":"7e0865d9-37e1-4f64-a1e9-cc8d221ebcdb","topicSidebar":"$L3a","topicSlug":"sequences-and-series","topicTitle":"3.4 Sequences and Series (A-level only)"}]}],"$L3b","$L3c"]}]

3.4 Sequences and Series (A-level only)

Question 18

3.4 Sequences and Series (A-level only) Questions

3.4 Sequences and Series (A-level only)

Question 18

3.4 Sequences and Series (A-level only) Questions