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Question
If a2, b2, c2 are in A.P., prove that \[\frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b}\] are in A.P.
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Solution
\[a^2 , b^2 , c^2 \text { are in A . P } . \]
\[ \therefore 2 b^2 = a^2 + c^2 \]
\[ \Rightarrow b^2 - a^2 = c^2 - b^2 \]
\[ \Rightarrow (b + a)(b - a) = (c - b)(c + b)\]
\[ \Rightarrow \frac{b - a}{c + b} = \frac{c - b}{b + a}\]
\[ \Rightarrow \frac{b - a}{(c + a)(c + b)} = \frac{c - b}{(b + a)(c + a)} \left[ \text { Multiplying both the sides by } \frac{1}{c + a} \right]\]
\[ \Rightarrow \frac{1}{c + a} - \frac{1}{b + c} = \frac{1}{a + b} - \frac{1}{c + a}\]
\[ \therefore ' \frac{1}{b + c}, \frac{1}{c + a}, \frac{1}{a + b} \text { are in A . P } . \]
\[\text { Multiplying each term by } (a + b + c): \]
\[\frac{a + b + c}{b + c}, \frac{a + b + c}{c + a}, \frac{a + b + c}{a + b} \text { are in A . P } . \]
\[\text { Thus }, \frac{a}{b + c} + 1 , \frac{b}{c + a} + 1 , \frac{c}{a + b} + 1 \text { are in A . P } . \]
\[\text { Hence }, \frac{a}{b + c}, \frac{b}{c + a}, \frac{c}{a + b} \text { are in A . P } .\]
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