Question

The sum of infinite terms of a G.P. $x$ and on squaring each term of it, the 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

The correct option is C

$\frac{x^2-y}{x^2+y}$

Explanation for correct option:

Given the sum of infinite terms of a G.P. $=x$

Let a be the first term and r be the common ratio,

The series is$a,ar,a{r}^2\dots$

$S=\frac{a}{(1-r)}=x\dots \left(i\right)$

$\therefore a=x(1-r)$

When the terms are squared, it becomes $a^2,a^2{r}^2,a^2{r}^4,....$

Sum$=\frac{a^2}{1-r^2}$

Substitute $a$ in above equation

$$\begin{gathered} \Rightarrow \frac{x^2{\left(1-r\right)}^2}{1-r^2}=y \\ \Rightarrow \frac{x^2{\left(1-r\right)}^2}{\left(1-r\right)\left(1+r\right)}=y \\ \Rightarrow \frac{x^2\left(1-r\right)}{\left(1+r\right)}=y \\ \Rightarrow \frac{x^2}{y}=\frac{1+r}{1-r} \end{gathered}$$

Applying componendo dividendo rule

$$\begin{gathered} \Rightarrow \frac{x^2-y}{x^2+y}=\frac{\left(1+r\right)-\left(1-r\right)}{\left(1+r\right)+\left(1-r\right)} \\ \Rightarrow \frac{x^2-y}{x^2+y}=\frac{2r}{2} \\ \Rightarrow \frac{x^2-y}{x^2+y}=r \end{gathered}$$

So $r=\frac{x^2-y}{x^2+y}$

Hence, option is C correct.