A geometric sequence has first term a, common ratio r and sum to infinity 76. A second geometric sequence has first term a, common ratio ${r^3}$ and sum to infinity 36.

Find r.

for the first series $\frac{a}{{1 - r}} = 76$     A1

for the second series $\frac{a}{{1 - {r^3}}} = 36$     A1

attempt to eliminate a e.g. $\frac{{76(1 - r)}}{{1 - {r^3}}} = 36$     M1

simplify and obtain $9{r^2} + 9r - 10 = 0$     (M1)A1

Note: Only award the M1 if a quadratic is seen.

obtain $r = \frac{{12}}{{18}}$ and $- \frac{{30}}{{18}}$     (A1)

$r = \frac{{12}}{{18}}\left( { = \frac{2}{3} = 0.666 \ldots } \right)$     A1

Note: Award A0 if the extra value of r is given in the final answer.

Total [7 marks]

Almost all candidates obtained the cubic equation satisfied by the common ratio of the first sequence, but few were able to find its roots. One of the roots was $r = 1$.