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Question
If a, b, c are in G.P., prove that:
\[\frac{1}{a^2 - b^2} + \frac{1}{b^2} = \frac{1}{b^2 - c^2}\]
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Solution
a, b and c are in G.P.
\[\therefore b^2 = ac\] .......(1)
\[\text { LHS } = \frac{1}{a^2 - b^2} + \frac{1}{b^2}\]
\[ = \frac{b^2 + a^2 - b^2}{\left( a^2 - b^2 \right) b^2}\]
\[ = \frac{a^2}{\left( a^2 b^2 - b^4 \right)}\]
\[ = \frac{a^2}{a^2 \left( ac \right) - \left( ac \right)^2}\]
\[ = \frac{1}{ac - c^2}\]
\[ = \frac{1}{b^2 - c^2} = \text { RHS }\]
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