# Math Help - Geometric progression

1. ## Geometric progression

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 ?

2. Originally Posted by thereddevils
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 ?
$S_6 = \frac{a - a r^6}{1 - r}$ and $S_3 = \frac{a - a r^3}{1 - r}$. It follows from $\frac{S_6}{S_3} = 9$ that $\frac{1 - r^6}{1 - r^3} = 9 \Rightarrow \frac{(1 - r^3)(1 + r^3)}{1 - r^3} = 9$ (nb: $r \neq 1$).

Therfore $1 + r^3 = 9$....

3. If you were to denote the first 6 terms of the geometric sequence as $a, ar, \ldots , ar^5$, then:

$\frac{a+ar+ar^2+\cdots+ar^5}{a+ar+ar^2} = 9$

$1 + \frac{ar^3+ar^4+ar^5}{a+ar+ar^2} = 9$

$1 + \frac{ar^3(1+r+r^2)}{a(1+r+r^2)} = 9$

$\ldots$