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If A, B, C, D Are in G.P., Prove That: 1 a 2 + B 2 , 1 B 2 − C 2 , 1 C 2 + D 2 Are in G . P . - Mathematics

If a, b, c, d are in G.P., prove that:

\[\frac{1}{a^2 + b^2}, \frac{1}{b^2 - c^2}, \frac{1}{c^2 + d^2} \text { are in G . P } .\]

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Solution

a, b, c and d are in G.P.

\[\therefore b^2 = ac\]

\[ad = bc \]

\[ c^2 = bd\]   .......(1)

\[\left( \frac{1}{b^2 + c^2} \right)^2 = \left( \frac{1}{b^2} \right)^2 + \frac{2}{b^2 c^2} + \left( \frac{1}{c^2} \right)^2 \]

\[ \Rightarrow \left( \frac{1}{b^2 + c^2} \right)^2 = \left( \frac{1}{ac} \right)^2 + \frac{1}{b^2 c^2} + \frac{1}{b^2 c^2} + \left( \frac{1}{bd} \right)^2 \left[ \text { Using } (1) \right]\]

\[ \Rightarrow \left( \frac{1}{b^2 + c^2} \right)^2 = \frac{1}{a^2 c^2} + \frac{1}{a^2 d^2} + \frac{1}{b^2 c^2} + \frac{1}{b^2 d^2} \left[ \text { Using  }(1) \right]\]

\[ \Rightarrow \left( \frac{1}{b^2 + c^2} \right)^2 = \frac{1}{a^2}\left( \frac{1}{c^2} + \frac{1}{d^2} \right) + \frac{1}{b^2}\left( \frac{1}{c^2} + \frac{1}{d^2} \right)\]

\[ \Rightarrow \left( \frac{1}{b^2 + c^2} \right)^2 = \left( \frac{1}{a^2 + b^2} \right)\left( \frac{1}{c^2} + \frac{1}{d^2} \right)\]

\[\text{ Therefore }, \left( \frac{1}{b^2 + c^2} \right), \left( \frac{1}{c^2 + d^2} \right)\text {  and } \left( \frac{1}{b^2 + c^2} \right) \text { are also in G . P } .\]

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Exercise 20.5 | Q 11.3 | Page 46
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