Question

If a, b, c are in G.P. then $\frac{1}{a^2-b^2}+\frac{1}{b^2}+\frac{1}{b^2-c^2}=0$

• A. True
• B. False

Answer 1

The correct option is B False

$a,b,c$ are in G.P

$\Rightarrow \frac{b}{a}=\frac{c}{b}$

$\Rightarrow b^2=ac$

$\frac{1}{a^2-b^2}+\frac{1}{b^2}+\frac{1}{b^2-c^2}$

$=\frac{1}{a^2-ac}+\frac{1}{ac}+\frac{1}{ac-c^2}$

$=\frac{ac-c^2+a^2-ac}{(a^2-ac)(ac-c^2)}+\frac{1}{ac}$

$=\frac{-c^2+a^2}{a(a-c)c(a-c)}+\frac{1}{ac}$

$=\frac{(a-c)(a+c)}{a(a-c)c(a-c)}+\frac{1}{ac}$

$=\frac{(a+c)}{ac(a-c)}+\frac{1}{ac}$

$=\frac{1}{ac}[\frac{(a+c)}{(a-c)}+1]$

$=\frac{1}{ac}\left[\frac{a+c+a-c}{(a-c)}\right]$

$=\frac{1}{ac}\left[\frac{2a}{(a-c)}\right]$

$=\frac{2}{c(a-c)}$

$=\frac{2}{(ac-c^2)}$

$=\frac{2}{(b^2-c^2)}$