If a, b, c, d are in G.P., prove that:
\[\frac{1}{a^2 + b^2}, \frac{1}{b^2 - c^2}, \frac{1}{c^2 + d^2} \text { are in G . P } .\]
Solution
a, b, c and d are in G.P.
\[\therefore b^2 = ac\]
\[ad = bc \]
\[ c^2 = bd\] .......(1)
\[\left( \frac{1}{b^2 + c^2} \right)^2 = \left( \frac{1}{b^2} \right)^2 + \frac{2}{b^2 c^2} + \left( \frac{1}{c^2} \right)^2 \]
\[ \Rightarrow \left( \frac{1}{b^2 + c^2} \right)^2 = \left( \frac{1}{ac} \right)^2 + \frac{1}{b^2 c^2} + \frac{1}{b^2 c^2} + \left( \frac{1}{bd} \right)^2 \left[ \text { Using } (1) \right]\]
\[ \Rightarrow \left( \frac{1}{b^2 + c^2} \right)^2 = \frac{1}{a^2 c^2} + \frac{1}{a^2 d^2} + \frac{1}{b^2 c^2} + \frac{1}{b^2 d^2} \left[ \text { Using }(1) \right]\]
\[ \Rightarrow \left( \frac{1}{b^2 + c^2} \right)^2 = \frac{1}{a^2}\left( \frac{1}{c^2} + \frac{1}{d^2} \right) + \frac{1}{b^2}\left( \frac{1}{c^2} + \frac{1}{d^2} \right)\]
\[ \Rightarrow \left( \frac{1}{b^2 + c^2} \right)^2 = \left( \frac{1}{a^2 + b^2} \right)\left( \frac{1}{c^2} + \frac{1}{d^2} \right)\]
\[\text{ Therefore }, \left( \frac{1}{b^2 + c^2} \right), \left( \frac{1}{c^2 + d^2} \right)\text { and } \left( \frac{1}{b^2 + c^2} \right) \text { are also in G . P } .\]
