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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.
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We have $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P. Adding 1 to each term $a\left(\frac{1}{b}+\frac{1}{c}\right)+1, b\left(\frac{1}{c}+\frac{1}{a}\right)+1, c\left(\frac{1}{a}+\frac{1}{b}\right)+1$ are in A.P. or $a\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right), c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$ are in A.P. Dividing by $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ we have $a, b, c$ are in A.P.
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