The sum of the second and third terms of a geometric sequence is 96.

The sum to infinity of this sequence is 500.

Find the possible values for the common ratio, $r$.

$ar + a{r^2} = 96$    A1

Note:     Award A1 for any valid equation involving $a$ and $r$, eg, $\frac{{a(1 - {r^3})}}{{1 - r}} - a = 96$.

$\frac{a}{{1 - r}} = 500$    A1

EITHER

attempting to eliminate $a$ to obtain $500r(1 - {r^2}) = 96$ (or equivalent in unsimplified form)     (M1)

OR

attempting to obtain $a = \frac{{96}}{{r + {r^2}}}$ and $a = 500(1 - r)$     (M1)

THEN

attempting to solve for $r$     (M1)

$r = 0.2{\text{ }}\left( { = \frac{1}{5}} \right)$ or $r = 0.885{\text{ }}\left( { = \frac{{\sqrt {97}  - 1}}{{10}}} \right)$     A1A1

[6 marks]

Reasonably well done. Quite a number of candidates included a solution outside $- 1 < r < 1$.