Question

If $a+b+c\ne 0$ and $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$ are in A. P., prove that $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are also in A.P.

Answer 1

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

Therefore,

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

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

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

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

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

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

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

$ac-bc=ab-ac$

$\frac{ac}{abc}-\frac{bc}{abc}=\frac{ab}{abc}-\frac{ac}{abc}$

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

Hence, $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in AP.

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