Question

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

Answer 1

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

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

$\frac{ac+a^2-b^2-bc}{ab}=\frac{ab+b^2-c^2-ac}{bc}$

$\frac{c(a-b)+(a^2-b^2)}{ab}=\frac{a(b-c)+(b^2-c^2)}{bc}$

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

$\frac{(a-b)}{ab}=\frac{(b-c)}{bc}$

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

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

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