Question

If $(b+c)(c+a)(a+b)$ are in H.P., then show that $\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}$ are also in H.P

Answer 1

Given that:

$(b+c),(c+a),(a+b)$ are in H.P.

To show:

$\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}$ are in H.P.

Solution:

$(b+c),(c+a),(a+b)$ are in H.P.

So, $\frac{1}{(c+a)}-\frac{1}{(b+c)}=\frac{1}{(a+b)}-\frac{1}{(c+a)}$

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

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

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

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

So, $a^2,b^2,c^2$ are in A.P.

Therefore, $\frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}$ are in H.P.