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

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Question

(c) We have $\frac{a}{{1 - r}} = x$ &hellip;..$(i)$

and $\frac{{{a^2}}}{{1 - {r^2}}} = \frac{a}{{1 - r}}.\frac{a}{{1 + r}} = y$ &hellip;..$(ii)$

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) We have $\frac{a}{{1 - r}} = x$ &hellip;..$(i)$

and $\frac{{{a^2}}}{{1 - {r^2}}} = \frac{a}{{1 - r}}.\frac{a}{{1 + r}} = y$ &hellip;..$(ii)$

$ \Rightarrow $ $y = x.\frac{a}{{1 + r}} = x.\frac{{x(1 - r)}}{{1 + r}}$

$ \Rightarrow $$\frac{y}{{{x^2}}} = \frac{{1 - r}}{{1 + r}}$

$ \Rightarrow $ $\frac{{{x^2}}}{y} = \frac{{1 + r}}{{1 - r}}$

$ \Rightarrow $$\frac{{{x^2}}}{y}(1 - r) = 1 + r$ 

$\Rightarrow r[1+ \frac{{x^2}}{{y}}] = -1 + \frac{{x^2}}{{y}}$

$\Rightarrow r=\frac{{{x^2} + y}}{{{x^2} - y}}$

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

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