Question

If $a,b,c$ are in HP ,then $\frac{1}{b-a}+\frac{1}{b-c}$ is equal to

• A. $\frac{2}{b}$
• B. $\frac{2}{a+c}$
• C. $\frac{1}{a}+\frac{1}{c}$
• D. none of these

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{2}{b}$
C $\frac{1}{a}+\frac{1}{c}$

Given $b=\frac{2ac}{a+c}$
$\therefore \frac{1}{b-a}+\frac{1}{b-c}=\frac{2b-(c+a)}{b^2-(a+c)b+ac}$
$=\frac{2b-\frac{2ac}{b}}{b^2-2ac+ac}$
$=2\frac{b^2-ac}{b(b^2-ac)}=\frac{2}{b}$
$=\frac{a+c}{ac}=\frac{1}{a}+\frac{1}{c}$
Hence, options A and C are correct