Question

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

Answer 1

Given:
$a\left(\frac{1}{b}+\frac{1}{c}\right),b\left(\frac{1}{c}+\frac{1}{a}\right),c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P.
$$\begin{gathered} \mathrm{By}\mathrm{adding}1\mathrm{to}\mathrm{each}\mathrm{term},\mathrm{we}\mathrm{get}: \\ a\left(\frac{1}{b}+\frac{1}{c}\right)+1,b\left(\frac{1}{c}+\frac{1}{a}\right)+1,c\left(\frac{1}{a}+\frac{1}{b}\right)+1\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \Rightarrow a\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right),b\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right),c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \mathrm{Dividing}\mathrm{all}\mathrm{terms}\mathrm{by}\frac{1}{a}+\frac{1}{b}+\frac{1}{c},\mathrm{we}\mathrm{get}: \\ \Rightarrow a\mathit{,}\mathit{}b\mathit{,}\mathit{}c\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}. \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$