Question

Given that $a^2,b^2,c^2$ are in an A.P. Prove that $\frac{1}{b+c},\frac{1}{a+c},\frac{1}{b+a}$ are in A.P

Answer 1

To prove $\frac{1}{b+c},\frac{1}{a+c},\frac{1}{b+a}$ are in A.P
Given that $a^2,b^2,c^2$ are in a A.P
Therefore,
$2b^2=a^2+c^2{b}^2+b^2=a^2+c^2{b}^2-a^2=c^2-b^2(b+a).(b-a)=(c+b)(c-b)\frac{(b-a)}{(c+b)}=\frac{(c-b)}{(b+a)}Nowdividingbothsidesby\frac{1}{(c+a)}\frac{(b-a)}{(c+b)(c+a)}=\frac{(c-b)}{(b+a)(c+a)}\frac{(b+c)-(c+a)}{(b+c)(c+a)}=\frac{(c+a)-(a+b)}{(a+b)(c+a)}\frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}\frac{2}{c+a}=\frac{1}{a+b}+\frac{1}{b+c}$
Therefore,
$\frac{1}{b+c},\frac{1}{c+a}$ and $\frac{1}{a+b}$ are in A.P.