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

Here,
$a(\frac{1}{b}+\frac{1}{c}),b(\frac{1}{c}+\frac{1}{a}),c(\frac{1}{a}+\frac{1}{b})$ are in A.P.
By adding 1 to each term, we get
$\Rightarrow a\left(\frac{c+b}{bc}\right)+1,b\left(\frac{a+c}{ca}\right)+1,c\left(\frac{b+a}{ab}\right)+1$ are in A.P.
$\Rightarrow \frac{ca+ab}{bc}+1,\frac{ab+bc}{ca}+1,\frac{bc+ca}{ab}+1$ are in A.P.
Dividing all terms by ab + bc + ca
$\frac{1}{bc},\frac{1}{ca},\frac{1}{ab}$ are in A.P.
Multiplying by abc
$\frac{abc}{bc},\frac{abc}{ca},\frac{abc}{ab}$ are in A.P.
$\Rightarrow a,b,c$ are in A.P.