Question

If $\frac{1}{a}+\frac{1}{a-2b}+\frac{1}{c}+\frac{1}{c-2b}=0$ and a, b, c are not in A.P., then

• A. a, b, c are in G.P.
• B. $a,\frac{b}{2},c$ are in A.P.
• C. $a,\frac{b}{2},c$ are in H.P.
• D. a, 2b, c are in H.P.

Answer 1

The correct option is D a, 2b, c are in H.P.

Given $\frac{1}{a}+\frac{1}{a-2b}+\frac{1}{c}+\frac{1}{c-2b}=0$
$\Rightarrow \left(\frac{1}{a}+\frac{1}{c-2b}\right)+\left(\frac{1}{c}+\frac{1}{a-2b}\right)=0$
$(a+c-2b)\left(\frac{1}{a(c-2b)}+\frac{1}{c(a-2b)}\right)=0$

As $a+c-2b\ne 0$

$\frac{1}{a(c-2b)}+\frac{1}{c(a-2b)}=0$

$\Rightarrow ac=bc+ab$

$\Rightarrow \frac{1}{b}=\frac{1}{a}+\frac{1}{c}$
$\Rightarrow \frac{2}{2b}=\frac{1}{a}+\frac{1}{c}$

i.e. $a,2b,c$ are in H.P.