The sum of infinite terms of a G.P. is x and | vedclass

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(C) Let the $G.P.$ be $a, ar, ar^2, \dots$ where $|r| < 1$.The sum of infinite terms is $x = \frac{a}{1 - r} \implies a = x(1 - r) \dots (i)$.When eac

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$\frac{x^2 - y}{x^2 + y}$

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(C) Let the $G.P.$ be $a, ar, ar^2, \dots$ where $|r| The sum of infinite terms is $x = \frac{a}{1 - r} \implies a = x(1 - r) \dots (i)$.When each term is squared,the new series is $a^2, a^2r^2, a^2r^4, \dots$,which is a $G.P.$ with first term $a^2$ and common ratio $r^2$.The sum of this new series is $y = \frac{a^2}{1 - r^2} = \frac{a^2}{(1 - r)(1 + r)} \dots (ii)$.Substituting $a = x(1 - r)$ from $(i)$ into $(ii)$:$y = \frac{[x(1 - r)]^2}{(1 - r)(1 + r)} = \frac{x^2(1 - r)^2}{(1 - r)(1 + r)} = \frac{x^2(1 - r)}{1 + r}$.Rearranging to solve for $r$:$y(1 + r) = x^2(1 - r)$$y + yr = x^2 - x^2r$$yr + x^2r = x^2 - y$$r(x^2 + y) = x^2 - y$$r = \frac{x^2 - y}{x^2 + y}$.

The sum of infinite terms of a $G.P.$ is $x$ and on squaring each term of it,the sum becomes $y$. Then the common ratio of this series is

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