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If B + C a , C + a B , a + B C Are in A.P., Prove That: 1 a , 1 B , 1 C Are in A.P.

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प्रश्न

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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उत्तर

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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अध्याय 19: Arithmetic Progression - Exercise 19.5 [पृष्ठ ४२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 19 Arithmetic Progression
Exercise 19.5 | Q 4.1 | पृष्ठ ४२

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