Question

The sum of first three terms of a G.P. is $16$ and the sum of its next three terms is $128.$ Then the sum of next three terms of G.P is

• A. $\frac{2008}{7}$
• B. $\frac{1604}{7}$
• C. $\frac{4016}{7}$
• D. $1024$

Answer 1

The correct option is D $1024$

Let $a$ be the first term and $r$ be the common ratio of the given G.P.
Then, $a+ar+a{r}^2=16$ and $a{r}^3+a{r}^4+a{r}^5=128$
$\Rightarrow a(1+r+r^2)=16\cdots \left(1\right)$ and
$a{r}^3(1+r+r^2)=128\cdots \left(2\right)$

On dividing $\left(2\right)$ by $\left(1\right),$ we get
$r^3=8=2^3\Rightarrow r=2$

Putting $r=2$ in $\left(1\right),$ we get
$a(1+2+4)=16\Rightarrow a=\frac{16}{7}$
Thus, in the given G.P, we have $a=\frac{16}{7}$ and $r=2>1$

So, the sum of next three terms is,
$S=S_9-16-128\Rightarrow S=\frac{a(r^9-1)}{(r-1)}-144\Rightarrow S=\frac{\frac{16}{7}\times 511}{1}-144\Rightarrow S=1024$