Question

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

Answer 1

It is given that $a^2,b^2,c^2$ are in A.P, then the common difference can be determined as follows:

$d_1=b^2-a^2$ and $d_2=c^2-b^2$

Since $a,b,c$ are in A.P, therefore the common difference must be equal that is:

$d_1=d_2\Rightarrow b^2-a^2=c^2-b^2\Rightarrow (b+a)(b-a)=(c+b)(c-b)\Rightarrow (b+a)(b-a+c-c)=(c+b)(c-b+a-a)$

$\Rightarrow (b+a)\left[\right(b+c)-(a+c\left)\right]=(c+b)\left[\right(c+a)-(b+a\left)\right]\Rightarrow (b+a)(b+c)-(b+a)(a+c)=(c+b)(c+a)-(c+b)(b+a)\Rightarrow \frac{1}{c+a}-\frac{1}{b+a}=\frac{1}{a+b}-\frac{1}{c+a}\left(Divideby\right(a+b\left)\right(b+c\left)\right(c+a\left)\right)$

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