Question

The sum of infinite terms of a G.P is $x$ and on squaring each term of it, the sum will be $y$, then the common ratio of this series is
A. $\frac{{{x}^{2}}-{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}$
B. $\frac{{{x}^{2}}+{{y}^{2}}}{{{x}^{2}}-{{y}^{2}}}$
C. $\frac{{{x}^{2}}-y}{{{x}^{2}}+y}$
D. $\frac{{{x}^{2}}+y}{{{x}^{2}}-y}$

Answer 1

Hint: In this question, we are to find the relationship between the given sums. Since the given sums are in G.P, we can use the sum of infinite terms formula and the required relation is calculated.

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 step by step solution:Consider a series of terms which 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 ${{S}_{\infty }}=x$.
So,
$x=\frac{a}{1-r}\text{ }...(1)$
On squaring each term in the 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 ${{S}_{\infty }}=y$.
So,
$\begin{align}
  & y=\frac{{{a}^{2}}}{1-{{r}^{2}}} \\
 & \text{ }=\frac{a}{1-r}\times \frac{a}{1+r} \\
\end{align}$
On substituting (1), we get
$y=x\left( \frac{a}{1+r} \right)$
Here $a=x(1-r)$ from (1).
So,
$y=x\left( \frac{x(1-r)}{1+r} \right)$
On simplifying,
$\begin{align}
  & y={{x}^{2}}\left( \frac{1-r}{1+r} \right) \\
 & \Rightarrow \frac{y}{{{x}^{2}}}=\frac{1-r}{1+r} \\
 & \Rightarrow \frac{{{x}^{2}}}{y}=\frac{1+r}{1-r} \\
 & \Rightarrow \frac{{{x}^{2}}}{y}(1-r)=1+r \\
\end{align}$
On simplifying,
$\begin{align}
  & \Rightarrow \frac{{{x}^{2}}}{y}-\frac{{{x}^{2}}}{y}(r)=1+r \\
 & \Rightarrow \frac{{{x}^{2}}}{y}-1=r+\frac{{{x}^{2}}}{y}(r) \\
 & \Rightarrow \frac{{{x}^{2}}}{y}-1=r(1+\frac{{{x}^{2}}}{y}) \\
 & \Rightarrow r=\frac{(\frac{{{x}^{2}}}{y}-1)}{(\frac{{{x}^{2}}}{y}+1)} \\
\end{align}$
\[\begin{align}
  & \Rightarrow r=\frac{(\frac{{{x}^{2}}-y}{y})}{(\frac{{{x}^{2}}+y}{y})} \\
 & \Rightarrow r=\frac{{{x}^{2}}-y}{{{x}^{2}}+y} \\
\end{align}\]


Option ‘C’ is correct

Note: Here the given sums are in G.P. So, using the sum of infinite terms formula, the required expression is framed.