Question

If the sum of first $6$ terms is $9$ times to the sum of first $3$ terms of the same G.P., then common ratio of the series will be

• A. $-2$
• B. $2$
• C. $1$
• D. $\frac{1}{2}$

Answer 1

The correct option is B

$2$

Explanation for correct option

Let $a$ be the first term and $r$ be the common ratio of the GP.

We know that the sum of the $n$ terms of GP, is $S_n=\frac{a\left(r^n-1\right)}{r-1}$

Given $S_6=9S_3$

$$\begin{gathered} \Rightarrow \frac{a\left(r^6-1\right)}{r-1}=9\frac{a\left(r^3-1\right)}{r-1} \\ \Rightarrow r^6-1=9\left(r^3-1\right) \\ \Rightarrow \left(r^3-1\right)\left(r^3+1\right)=9\left(r^3-1\right)\left[\because a^2-b^2=\left(a+b\right)\left(a-b\right)\right] \\ \Rightarrow \left(r^3+1\right)=9 \\ \Rightarrow r^3=8 \\ \Rightarrow r=2 \end{gathered}$$

Hence, option (B) i.e. $2$ is correct