Question

If $\frac{1}{a(b+c)},\frac{1}{b(c+a)},\frac{1}{c(a+b)}$ be in HP, then a,b,c are in

• A. G.P
• B. A.P
• C. A.G.P
• D. H.P

Answer 1

The correct option is D H.P

$\frac{1}{a(b+c)},\frac{1}{b(c+a)},\frac{1}{c(a+b)}$ are in H.P.

$\Rightarrow a(b+c),b(c+a),c(a+b)$ are in A.P.

Subtract $ab+bc+ca$ from each term.

$\Rightarrow -bc,-ca,-ab$ are in A.P.

Divide each term with $-abc$,

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

$\therefore a,b,c$ are in H.P.

Hence, option D.