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 \(r = 1\).