Question

If a, b, c are in H.P, then the value of
$(\frac{1}{b}+\frac{1}{c}-\frac{1}{a})(\frac{1}{c}+\frac{1}{a}-\frac{1}{b})$, is

Answer 1

Given,

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

$\frac{1}{a}+\frac{1}{c}=\frac{2}{b}$............(1)

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

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

$={\left(\frac{1}{b}\right)}^2-{(\frac{1}{a}-\frac{1}{c})}^2$

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

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

$=\frac{1}{b^2}-[(\frac{1}{a^2}+\frac{1}{c^2})-\frac{4}{ac}]$

From (1)

$=\frac{1}{b^2}-[{\left(\frac{2}{b}\right)}^2-\frac{4}{ac}]$

$=\frac{1}{b^2}-\frac{4}{b^2}+\frac{4}{ac}$

$=\frac{4}{ac}-\frac{3}{b^2}$