easy problem

Let $a$ be the first term of the G.P. and $r$ the ratio. In general, for a G.P. the sum $S_{n}$ of the first $n$ terms is

$S_{n} = \dfrac{a(r^{n} - 1)}{r - 1}$ .

We are given that

$\dfrac{S_{6}}{S_{3}} = 9 \Longrightarrow \dfrac{\dfrac{a(r^{6} - 1)}{r - 1}}{\dfrac{a(r^{3} - 1)}{r - 1}} = 9 \Longrightarrow r^{3} + 1 = 9$ ,

since $r^{6} - 1 = (r^{3} + 1)(r^{3} - 1)$ . Thus $r^{3} = 8 \Longrightarrow \boxed{r = 2}$ .

(Note that we could rule out the possibility of $r = 1$ since that would have made $\frac{S_{6}}{S_{3}} = 2$ and not $9$ .)