Question

If $abc$ are in H.P. then prove that $\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}$

Answer 1

a, b, c are in H.P

$\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$2ac=b(a+c)$

Add $2b^2$ on both sides

$2ac+2b^2=b(a+c)+2b^2$

Add $-2b(a+c)$ on both sides

$2b^2-2b(a+c)+2ac=-b(a+c)+2b^2$

$2(b^2-b(a+c)+ac)=b(2b-(a+c\left)\right)$

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

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

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

$\frac{1}{a}+\frac{1}{c}=\frac{1}{b-a}+\frac{1}{b-c}$.