Question

Sum of infinite number of terms in G.P. is $29$ and sum of their square is $100$. The common ratio of G.P. is

• A. $5$
• B. $\frac{3}{5}$
• C. $\frac{8}{5}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B

$\frac{3}{5}$

Explanation for the correct option :

Step 1 : Using the sum formula of an infinite G.P.

Let $a,ar,a{r}^2,...$ be the G.P. series where $a$ is initial term and $r$ is common difference.

Formula to be used : We know that the the sum of the terms of an infinite G.P. with the initial term $a$ and common ratio $r$ is $\frac{a}{1-r}$.

Given that the sum of the G.P. is $20$. So, we must get

$\frac{\mathrm{a}}{1-\mathrm{r}}=20$ $\left(1\right)$

Now, squaring each term of the series, we get the series with the following terms :

$a^2,a^2{r}^2,a^2{r}^4,...$

This is again an infinite G.P. with initial term $a^2$ and common difference $r^2$.

Therefore, the sum of infinite terms of this G.P. is $\frac{a^2}{1-r^2}$.

But sum is given as $100$. Hence, we get

$\frac{a^2}{1-r^2}=100$ $\left(2\right)$

Step 2 : Solving equations $\left(1\right)$ and $\left(2\right)$ to find the value of $r$

The two equations obtained are :

$\frac{\mathrm{a}}{1-\mathrm{r}}=20$ $\left(1\right)$

$\frac{a^2}{1-r^2}=100$ $\left(2\right)$

Squaring both sides of equation $\left(1\right)$, we get :

$\frac{{\mathrm{a}}^2}{{(1-\mathrm{r})}^2}=400$ $\left(3\right)$

Divide equation $\left(3\right)$ by equation $\left(2\right)$, we obtain :

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{a^2}{{(1-r)}^2}}}{{\displaystyle \frac{a^2}{1-r^2}}}=\frac{400}{100} \\ \Rightarrow \frac{1-r^2}{{(1-r)}^2}=4 \\ \Rightarrow \frac{1+r}{1-r}=4 \\ \Rightarrow 1+r=4-4r \\ \Rightarrow 5r=3 \\ \Rightarrow r=\frac{3}{5} \end{gathered}$$

Therefore, option (B) is the correct option.