Sequence And Series Question 268
Question: If $ G $ be the geometric mean of $ x $ and $ y $ , then $ \frac{1}{G^{2}-x^{2}}+\frac{1}{G^{2}-y^{2}}= $
Options:
A) $ G^{2} $
B) $ \frac{1}{G^{2}} $
C) $ \frac{2}{G^{2}} $
D) $ 3G^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
As given $ G=\sqrt{xy} $
$ \therefore $ $ \frac{1}{G^{2}-x^{2}}+\frac{1}{G^{2}-y^{2}}=\frac{1}{xy-x^{2}}+\frac{1}{xy-y^{2}} $ $ =\frac{1}{x-y}{ -\frac{1}{x}+\frac{1}{y} }=\frac{1}{xy}=\frac{1}{G^{2}} $ .
BETA
