question_answer

The sum of infinite terms of a G.P. is  and on squaring the each term of it, the sum will be , then the common ratio of this series is [RPET 1988]

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) \[\frac{{{x}^{2}}+y}{{{x}^{2}}-y}\]

Correct Answer: C

Solution :

We have \[\frac{a}{1-r}=x\] ?..(i)  and  \[\frac{{{a}^{2}}}{1-{{r}^{2}}}=\frac{a}{1-r}.\frac{a}{1+r}=y\] ?..(ii)  \[y=x.\frac{a}{1+r}=x.\frac{x(1-r)}{1+r}\]\[\frac{y}{{{x}^{2}}}=\frac{1-r}{1+r}\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{y}=\frac{1+r}{1-r}\]\[\Rightarrow \]\[\frac{{{x}^{2}}}{y}(1-r)=1+r\] \[\Rightarrow \] \[1+(1+x)+(1+x+{{x}^{2}})+...\]\[\Rightarrow \]\[{{(c-a)}^{2}}=(a+c)(a-2b+c)\].