Sequence And Series Question 579
The sum of infinite terms of a G.P. is and on squaring each term of it, the sum will be , then the common ratio of this series is
[RPET 1988]
Options:
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} $
Show Answer
Answer:
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-2c+a) $
BETA
