Question

If the third term of a G.P. is 12 and its sixth term is 96, find the sum of 9 terms.

Answer 1

Given that: $T_3=12$ and $T_6=96$
$T_3=12$
$\Rightarrow a{r}^2=12$ ......... (1)
$T_6=96$
$a{r}^5=96$ ........ (2)
divide (2) by (1): $\frac{a{r}^5}{a{r}^2}=\frac{96}{12}\therefore r^3=8$
$r^3=2^3$
$\therefore r=2$
Consider, $a{r}^2=12$
$a(2{)}^2=12\therefore a\times 4=12\therefore a=3$
Now, let us find $S_9$.
$S_n=\frac{a(r^n-1)}{r-1}$
$S_9=\frac{3(2^9-1)}{2-1}=\frac{3(512-1)}{1}=3\left(511\right)$
$\therefore S_9=1533$