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Question
If \[\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}\] are in A.P., prove that:
\[\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\] are in A.P.
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Solution
Since
\[\frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c}\] are in A.P., we have:
\[\frac{c + a}{b} - \frac{b + c}{a} = \frac{a + b}{c} - \frac{c + a}{b}\]
\[ \Rightarrow \frac{ac + a^2 - b^2 - bc}{ab} = \frac{ab + b^2 - c^2 - ac}{bc}\]
\[ \Rightarrow \frac{\left( a + b \right)\left( a - b \right) + c\left( a - b \right)}{ab} = \frac{\left( b + c \right)\left( b - c \right) + a\left( b - c \right)}{bc}\]
\[ \Rightarrow \frac{\left( a - b \right)\left( a + b + c \right)}{ab} = \frac{\left( b - c \right)\left( a + b + c \right)}{bc}\]
\[ \Rightarrow \frac{\left( a - b \right)}{ab} = \frac{\left( b - c \right)}{bc}\]
\[ \Rightarrow \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b}\]
Hence,
\[\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\] are in A.P.
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