Question

If $a,b,c$ are in harmonic progression, prove:
$\frac{b+a}{b-a}+\frac{b+c}{b-c}=2$

Answer 1

Given,

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

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

$b+a=\frac{2ac}{a+c}+a=\frac{a^2+3ac}{a+c}$

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

$b+c=\frac{c^2+3ac}{a+c}$

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

$\frac{b+a}{b-a}+\frac{b+c}{b-c}$

$=\frac{\frac{a^2+3ac}{a+c}}{\frac{ac-a^2}{a+c}}+\frac{\frac{c^2+3ac}{a+c}}{\frac{ac-c^2}{a+c}}$

$=\frac{a(a+3c)}{a(c-a)}+\frac{c(c+3a)}{c(a-c)}$

$=\frac{a+3c}{c-a}-\frac{c+3a}{c-a}$

$=\frac{a+3c-c-3a}{c-a}$

$=2$

Hence proved.