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Question
If \[\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\] are in A.P., prove that:
\[\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}\] are in A.P.
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Solution
\[\text { Given }: \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text { are in A . P } . \]
\[ \therefore \frac{2}{b} = \frac{1}{a} + \frac{1}{c}\]
\[ \Rightarrow 2ac = ab + bc . . . . (1)\]
\[\text { To prove }: \frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c} \text { are in A . P } . \]
\[ 2\left( \frac{a + c}{b} \right) = \frac{b + c}{a} + \frac{a + b}{c}\]
\[ \Rightarrow 2ac(a + c) = bc(b + c) + ab(a + b)\]
\[\text { LHS: }2ac(a + c)\]
\[ = (ab + bc)(a + c) (\text { From }(1))\]
\[ = a^2 b + 2abc + b c^2 \]
\[\text { RHS: } bc(b + c) + ab(a + b)\]
\[ = b^2 c + b c^2 + a^2 b + a b^2 \]
\[ = b^2 c + a b^2 + b c^2 + a^2 b\]
\[ = b(bc + ab) + b c^2 + a^2 b\]
\[ = 2abc + b c^2 + a^2 b \]
\[ = a^2 b + 2abc + b c^2 (\text { From }(1))\]
\[ \therefore \text { LHS = RHS }\]
\[\text { Hence, proved } . \]
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