The first three terms of a geometric series are (3k+3)(3k+3)(3k+3), (k+3)(k+3)(k+3), and k k\,k respectively, where k k\,k is a positive constant.
Show that 2k2−3k−9=02k^2 - 3k - 9 = 02k2−3k−9=0.
Hence show that k=3k = 3k=3.
Find the common ratio.
Find the sum to infinity of the series.
231 exam-style questions on WJEC A Level Maths 3.4 Sequences and Series (A-level only), covering 3.4.1 Sequences and Series (A-level only), 3.4.2 Sequences and Series (A-level only), 3.4.3 Sequences and Series (A-level only), 3.4.4 Sequences and Series (A-level only), 3.4.5 Sequences and Series (A-level only), 3.4.6 Sequences and Series (A-level only), and 3.4.7 Sequences and Series (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.