Question

If G be the GM between $x$ and $y$, then the value of $\frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}$ is equal to.

• A. $G^2$
• B. $\frac{2}{G^2}$
• C. $\frac{1}{G^2}$
• D. $3G^2$

Answer 1

The correct option is C $\frac{1}{G^2}$

Given, $G^2=xy..\left(1\right)$

$\therefore \frac{1}{G^2-x^2}+\frac{1}{G^2-y^2}=\frac{1}{xy-x^2}+\frac{1}{xy-y^2}$

$=\frac{1}{y-x}(\frac{1}{x}-\frac{1}{y})=\frac{1}{xy}=\frac{1}{G^2}$
Hence, option 'C' is correct.