Question

If the sum of an infinite G.P and the sum of squares of its terms is 3, then the common ratio of the first series is
A. $1$
B. $\frac{1}{2}$
C. $\frac{2}{3}$
D. $\frac{3}{2}$

Answer 1

Hint: In this question, we are to find the common ratio of the first series. Since the sums are equal for the given series, the required common ratio is obtained by equating both sums.

Formula Used:The sum of the infinite terms in G.P series is calculated by
 ${{S}_{\infty }}=\frac{a}{1-r}$ where $r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{\infty }}$ is the sum of the infinite terms of the series; $a$ is the first term in the series, and $r$ is the common ratio.



Complete answer:Consider a series of terms that are in G.P as
$a,ar,a{{r}^{2}},.....\infty $
Thus, the sum of the infinite terms of the series is calculated by the formula,
${{S}_{\infty }}=\frac{a}{1-r}$
On substituting,
\[{{S}_{\infty }}=\frac{a}{1-r}\]
But it is given that ${{S}_{\infty }}=3$.
So,
$\begin{align}
  & 3=\frac{a}{1-r} \\
 & \Rightarrow a=3(1-r)\text{ }...(1) \\
\end{align}$
On squaring each term in the first series, we get
${{a}^{2}},{{(ar)}^{2}},{{(a{{r}^{2}})}^{2}},.....\infty $
Since the series is a geometric series, the common ratio is
$r=\frac{{{a}_{n}}}{{{a}_{n-1}}}$
\[\begin{align}
  & \Rightarrow r=\frac{{{a}_{2}}}{{{a}_{1}}} \\
 & \text{ =}\frac{{{a}^{2}}{{r}^{2}}}{{{a}^{2}}} \\
 & \text{ }={{r}^{2}} \\
\end{align}\]
Thus, the sum of the infinite terms of the series is calculated by the formula,
${{S}_{\infty }}=\frac{a}{1-r}$
On substituting $a={{a}^{2}};r={{r}^{2}}$, we get
\[{{S}_{\infty }}=\frac{{{a}^{2}}}{1-{{r}^{2}}}\]
But it is given that ${{S}_{\infty }}=3$.
So,
$\begin{align}
  & 3=\frac{{{a}^{2}}}{1-{{r}^{2}}} \\
 & \Rightarrow {{a}^{2}}=3(1-{{r}^{2}}) \\
\end{align}$
On substituting (1), we get
$\begin{align}
  & {{\left( 3(1-r) \right)}^{2}}=3(1-{{r}^{2}}) \\
 & \Rightarrow 9{{(1-r)}^{2}}=3(1-r)(1+r) \\
 & \Rightarrow 3(1-r)(1-r)=(1-r)(1+r) \\
 & \Rightarrow 3(1-r)=(1+r) \\
\end{align}$
On simplifying,
$\begin{align}
  & \Rightarrow 3-3r=1+r \\
 & \Rightarrow 4r=2 \\
 & \Rightarrow r=\frac{1}{2} \\
\end{align}$
Option ‘B’ is correct
Note: