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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.
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Solution
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.
\[\text { By adding 1 to each term, we get }: \]
\[ a\left( \frac{1}{b} + \frac{1}{c} \right) + 1, b\left( \frac{1}{c} + \frac{1}{a} \right) + 1, c\left( \frac{1}{a} + \frac{1}{b} \right) + 1 \text { are in A . P } . \]
\[ \Rightarrow a\left( \frac{1}{b} + \frac{1}{c} + \frac{1}{a} \right), b\left( \frac{1}{c} + \frac{1}{a} + \frac{1}{b} \right), c\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \text { are in A . P } . \]
\[\text { Dividing all terms by } \frac{1}{a} + \frac{1}{b} + \frac{1}{c}, \text { we get }: \]
\[ \Rightarrow \text { a, b, c are in A . P } . \]
\[\text { Hence, proved } .\]
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