Question

If $a^2,b^2,c^2$ are in A.P. then $b+c,c+a,a+b$ are in H.P.

Answer 1

It is given that $a^2,b^2,c^2$ are in AP.

So,

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

$b^2-a^2=c^2-b^2$

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

Now diving both the sides by $\frac{1}{c+a}$

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

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

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

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

Therefore, $b+c,c+a,a+b$ are in HP.