मराठी

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

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प्रश्न

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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उत्तर

\[\text { Given }: \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \text { are  in A . P } . \]

\[ \therefore \frac{2}{b} = \frac{1}{a} + \frac{1}{c}\]

\[ \Rightarrow 2ac = ab + bc . . . . (1)\]

\[\text { To prove }: \frac{b + c}{a}, \frac{c + a}{b}, \frac{a + b}{c} \text { are in A . P } . \]

\[ 2\left( \frac{a + c}{b} \right) = \frac{b + c}{a} + \frac{a + b}{c}\]

\[ \Rightarrow 2ac(a + c) = bc(b + c) + ab(a + b)\]

\[\text { LHS: }2ac(a + c)\]

\[ = (ab + bc)(a + c) (\text { From }(1))\]

\[ = a^2 b + 2abc + b c^2 \]

\[\text { RHS: } bc(b + c) + ab(a + b)\]

\[ = b^2 c + b c^2 + a^2 b + a b^2 \]

\[ = b^2 c + a b^2 + b c^2 + a^2 b\]

\[ = b(bc + ab) + b c^2 + a^2 b\]

\[ = 2abc + b c^2 + a^2 b \]

\[ = a^2 b + 2abc + b c^2 (\text { From }(1))\]

\[ \therefore \text { LHS = RHS }\]

\[\text { Hence, proved } . \]

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

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 19 Arithmetic Progression
Exercise 19.5 | Q 1.1 | पृष्ठ ४२

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