Question

If $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in A.P., prove that:

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

Answer 1

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

By the definition of A.P. difference of two consecutive terms is same.

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

$\Rightarrow \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$

$\Rightarrow \frac{2}{b}=\frac{a+c}{ac}$

$\Rightarrow 2ac=ab+bc\dots \left(i\right)$

Let $a(b+c),b(c+a),c(a+b)$ be in an A.P.

By the definition of A.P. we may write as:

$2b(c+a)=a(b+c)+c(a+b)$

$R.H.S.=a(b+c)+c(a+b)$

$=ab+ac+ac+bc$

$=ab+2ac+bc$

$=ab+(ab+bc)+bc$ (from eq. (i))

$=2ab+2bc$

$=2b(c+a)=L.H.S$

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

Hence proved.