Question

If $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in A.P then prove that $a^2,b^2,c^2$ are in A.P

Answer 1

If $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ in A.P.

Prove that $a^2,b^2,c^2$ in A.P.

$\to \frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ in A.P.

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

$\therefore \frac{2}{(c+a)}=\frac{(b+c)+(a+b)}{(a+b)(b+c)}$

$\therefore 2(a+b)(b+c)=(a+2b+c)(c+a)$

$\therefore 2(ab+ac+b^2+bc)=(ac+2bc+c^2+a^2+2ab+ac)$

$\therefore 2ab+2ac+2b^2+2bc=2ac+2bc+a^2+c^2+2ab$

$\therefore 2b^2=a^2+c^2$

hence it is prove $a^2,c^2$ & $c^2$ also in A.P.