Question

If $a,b,c$ are in H.P ,then prove that $\frac{b+a}{b-a}+\frac{b+c}{b-c}$ is equal to $2$.

Answer 1

a,b,c are in H.P$\Rightarrow \frac{1}{a},\frac{1}{b},\frac{1}{c}areinA.P\Rightarrow \frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Rightarrow a+c=\frac{2ac}{b}\Rightarrow ab+bc=2acNow,\frac{b+a}{b-a}+\frac{b+c}{b-c}=\frac{(b+a)(b-c)+(b+c)(b-a)}{(b-a)(b-c)}=\frac{b^2+ab-bc-ac+b^2-ab+bc-ab}{b^2-ab+ac-bc}=\frac{2b^2-2ac}{b^2-2ac+ac}=\frac{2(b^2-ac)}{b^2-ac}=2$