Browse Questions

If x , y , z are in GP then , $\;\large\frac{1}{x^2-y^2}+\frac{1}{y^2}\;$ equals

$(a)\;\large\frac{1}{z^2-y^2}\qquad(b)\;\large\frac{1}{y^2-z^2}\qquad(c)\;\large\frac{1}{z^2-x^2}\qquad(d)\;\large\frac{1}{y^2-x^2}$

Answer : (b) $\;\large\frac{1}{y^2-z^2}$
Explanation : $\;\large\frac{1}{x^2-y^2}+\frac{1}{y^2}$
$=\frac{1}{x^2}(\large\frac{1}{1-r^2}+\frac{1}{r^2})$
$=\frac{1}{x^2}(\frac{1}{r^2}*\large\frac{1}{1-r^2})$
$=\large\frac{1}{y^2-z^2}\;.$