Question

If $a^2,b^2,c^2$ are in arithmetic progression, then the terms $\frac{1}{(a+b)},$ $\frac{1}{(c+a)},\frac{1}{(b+c)}$ will form

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is A A.P.

Given, $a^2,b^2,c^2$ are in AP.

$\therefore b^2-a^2=c^2-b^2$
$\frac{b-a}{b+c}=\frac{c-b}{a+b}$

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

$\frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}$

Hence $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in AP