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(B) Let the first series be $a + ar + ar^2 + \dots$ where $|r| < 1$.Then the sum is $S_1 = \frac{a}{1-r} = 3$,which implies $a = 3(1-r)$.The second se

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$\frac{1}{2}$

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(B) Let the first series be $a + ar + ar^2 + \dots$ where $|r| Then the sum is $S_1 = \frac{a}{1-r} = 3$,which implies $a = 3(1-r)$.The second series is $a^2 + a^2r^2 + a^2r^4 + \dots$,which is also a $G.P.$ with first term $a^2$ and common ratio $r^2$.The sum is $S_2 = \frac{a^2}{1-r^2} = 3$.Substituting $a = 3(1-r)$ into the second equation:$\frac{[3(1-r)]^2}{1-r^2} = 3$$\frac{9(1-r)^2}{(1-r)(1+r)} = 3$$\frac{3(1-r)}{1+r} = 1$$3 - 3r = 1 + r$$4r = 2$$r = \frac{1}{2}$.

If the sum of an infinite $G.P.$ and the sum of the squares of its terms is $3$,then the common ratio of the first series is

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