Question

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

• A. A.P
• B. G.P
• C. A.G.P
• D. H.P

Answer 1

The correct option is A A.P

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

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

$\Rightarrow (b-a)(b+a)=(c-b)(c+b)$

$\Rightarrow (b+c-c-a)(b+a)=(c+a-a-b)(c+b)$

$\Rightarrow [(b+c)-(c+a)](b+a)=[(c+a)-(a+b)](c+b)$

$\Rightarrow (a+b)(b+c)-(a+b)(c+a)=(b+c)(c+a)-(b+c)(a+b)$

Dividing both sides by $(a+b)(b+c)(c+a)$ we get

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

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