Question

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

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (B)

Clearly, $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P.
Let common difference of A.P. be d.
$\therefore \text{ Given expression }=\frac{\frac{1}{a}+\frac{1}{b}}{\frac{1}{a}-\frac{1}{b}}+\frac{\frac{1}{c}+\frac{1}{b}}{\frac{1}{c}-\frac{1}{b}}=\frac{\frac{1}{a}+\frac{1}{b}}{-d}+\frac{\frac{1}{c}+\frac{1}{b}}{d}$
$=\frac{-\frac{1}{a}-\frac{1}{b}+\frac{1}{c}+\frac{1}{b}}{d}=\frac{\frac{1}{c}-\frac{1}{a}}{d}=\frac{2d}{d}=2$