Question

If $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P., then prove that:
$\left(ii\right)$ $bc,ca,ab$ are in A.P.,

Answer 1

Given : $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A.P.
By the defination of A.P. difference of two consecative terms is same.
$\frac{c+a}{b}-\frac{b+c}{a}=\frac{a+b}{c}-\frac{c+a}{b}$
$\Rightarrow \frac{ac+a^2-b^2-bc}{ab}=\frac{ab+b^2-c^2-ac}{bc}$
$\Rightarrow \frac{(a+b)(b-c)+c(a-b)}{ab}=\frac{(b+c)(b-c)+a(b-c)}{bc}$
$\Rightarrow \frac{(a-b)(a+b+c)}{ab}=\frac{(b-c)(b+c+a)}{bc}$
$\Rightarrow \frac{(a-b)}{ab}=\frac{(b-c)}{bc}$
$\Rightarrow (a-b)c=a(b-c)$
$\Rightarrow ac-bc=ab-ac$
$\Rightarrow 2ac=ab+bc$
So, $bc,ca,ab$ are in A.P.
Hence proved