If the ratio of the sum of the first 6 terms of a G.P to the sum of the first 3 terms of the G.P is 9 , what is the common ratio of the G.P ?
$\displaystyle S_6 = \frac{a - a r^6}{1 - r}$ and $\displaystyle S_3 = \frac{a - a r^3}{1 - r}$. It follows from $\displaystyle \frac{S_6}{S_3} = 9$ that $\displaystyle \frac{1 - r^6}{1 - r^3} = 9 \Rightarrow \frac{(1 - r^3)(1 + r^3)}{1 - r^3} = 9$ (nb: $\displaystyle r \neq 1$).
Therfore $\displaystyle 1 + r^3 = 9 $....
If you were to denote the first 6 terms of the geometric sequence as $\displaystyle a, ar, \ldots , ar^5$, then:
$\displaystyle \frac{a+ar+ar^2+\cdots+ar^5}{a+ar+ar^2} = 9$
$\displaystyle 1 + \frac{ar^3+ar^4+ar^5}{a+ar+ar^2} = 9$
$\displaystyle 1 + \frac{ar^3(1+r+r^2)}{a(1+r+r^2)} = 9$
$\displaystyle \ldots$