Find the common ratio of the geometric sequence.

An arithmetic sequence ${v_1}$ , ${v_2}$ , ${v_3}$ , $...$ is such that ${v_2} = {u_2}$ and ${v_4} = {u_4}$ .

Find the greatest value of $N$ such that $\sum\limits_{n = 1}^N {{v_n}}  > 0$ .

${u_1} = 27$
$\frac{{81}}{2} = \frac{{27}}{{1 - r}}$     M1
$r = \frac{1}{3}$     A1

[2 marks]

${v_2} = 9$
${v_4} = 1$
$2d = - 8 \Rightarrow d = - 4$     (A1)
${v_1} = 13$     (A1)

$\frac{N}{2}\left( {2 \times 13 - 4\left( {N - 1} \right)} \right) > 0$     (accept equality)     M1

$\frac{N}{2}\left( {30 - 4N} \right) > 0$

$N\left( {15 - 2N} \right) > 0$

$N < 7.5$     (M1)

$N = 7$     A1

Note: $13 + 9 + 5 + 1 - 3 - 7 - 11 > 0 \Rightarrow N = 7$ or equivalent receives full marks.

[5 marks]

Part (a) was well done by most candidates. However (b) caused difficulty to most candidates. Although a number of different approaches were seen, just a small number of candidates obtained full marks for this question.

Part (a) was well done by most candidates. However (b) caused difficulty to most candidates. Although a number of different approaches were seen, just a small number of candidates obtained full marks for this question.