Question

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

• A. 1
• B. 2
• C. 3
• D. 1/2

Answer 1

Answer: (B)

Put $b=\frac{2ac}{a+c}$
$\Rightarrow \frac{b+a}{b-a}+\frac{b+c}{b-c}=\frac{\frac{2ac}{a+c}+a}{\frac{2ac}{a+c}-a}+\frac{\frac{2ac}{a+c}+c}{\frac{2ac}{a+c}-c}$
$=\frac{a+3c}{c-a}+\frac{3a+c}{a-c}=\frac{a+3c-3a-c}{c-a}=\frac{2(c-a)}{c-a}=2$