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 $|r| < 1$.Given the sum of infinite terms is $x = \frac{a}{1-r} \dots (i)$.Squaring each term gives t

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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| Given the sum of infinite terms is $x = \frac{a}{1-r} \dots (i)$.Squaring each term gives the series $a^2, a^2r^2, a^2r^4, \dots$,which is a $G.P.$ with first term $a^2$ and common ratio $r^2$.The sum of this new series is $y = \frac{a^2}{1-r^2} = \frac{a^2}{(1-r)(1+r)} \dots (ii)$.From $(i)$,$a = x(1-r)$. Substituting this 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 for $r$:$y(1+r) = x^2(1-r)$$y + yr = x^2 - x^2r$$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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