if a, b, c are in G.P., prove that $$\dfrac{ 1}{(a^2 – b^2)} + \dfrac{1}{b^2} = \dfrac{1}{(b^2 – c^2)}$$

Asked by Aaryan | 1 year ago |  70

##### Solution :-

Given that a, b, c are in GP.

By using the property of geometric mean,

b2 = ac

Let us consider LHS: $$\dfrac{ 1}{(a^2 – b^2) }+ \dfrac{1}{b^2}$$

Let us take LCM

= $$\dfrac{ (b^2 + a^2 – b^2)}{(a^2 – b^2)b^2}$$

= $$\dfrac{ a^2 }{ (a^2b^2 – b^4)}$$

= $$\dfrac{a^2 }{ a^2(b^2 – c^2)}$$

= $$\dfrac{1} {(b^2 – c^2)}$$

= RHS

LHS = RHS

Hence proved.

Answered by Aaryan | 1 year ago

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