Question

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

• A. A.P
• B. G.P
• C. H.P
• D. None of these

Answer 1

The correct option is C

H.P

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

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

⇒ (a - b)(a + b) = (c - a)(c + a)

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

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

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

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