Question

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

Answer 1

We have:

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

$\Rightarrow a\left(\frac{b+c}{bc}\right),b\left(\frac{a+c}{ac}\right),c\left(\frac{a+b}{ab}\right)$ are in A.P.

$\Rightarrow \frac{ab+ac}{bc},\frac{ab+bc}{ac},\frac{ac+bc}{ab}$ are in A.P.

$\Rightarrow \frac{ab+ac}{bc}+1,\frac{ab+bc}{ac}+1,\frac{ac+bc}{ab}+1$ are in A.P.

[Adding 1 to each number]

$\Rightarrow \frac{ab+ac+bc}{bc},\frac{ab+bc+ac}{ac},\frac{ac+bc+ab}{ab}$ are in A.P.

or $\frac{1}{bc},\frac{1}{ac},\frac{1}{ab}$ are in A.P.

[Dividing each number by ab + bc + ac]

or $\frac{abc}{bc},\frac{abc}{ac},\frac{abc}{ab}$ are in A.P.

[Multiplying each number by abc]

or a, b, c are in A.P.