If the sum of an infinite geometric series is | vedclass

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(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r} = 3$. This implies $

Vedclass · Accepted Answer

$\frac{3}{2}, \frac{1}{2}$

Vedclass · Answer

(B) Let the first term be $a$ and the common ratio be $r$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r} = 3$. This implies $a = 3(1-r)$. The squares of the terms form a new geometric series: $a^2, a^2r^2, a^2r^4, \dots$ The sum of this series is $S' = \frac{a^2}{1-r^2} = 3$. Substituting $a = 3(1-r)$ into the second equation: $\frac{[3(1-r)]^2}{1-r^2} = 3$ $\frac{9(1-r)^2}{(1-r)(1+r)} = 3$ $\frac{3(1-r)}{1+r} = 1$ $3 - 3r = 1 + r$ $4r = 2 \implies r = \frac{1}{2}$. Now,$a = 3(1 - \frac{1}{2}) = 3(\frac{1}{2}) = \frac{3}{2}$. Thus,the first term is $\frac{3}{2}$ and the common ratio is $\frac{1}{2}$.

If the sum of an infinite geometric series is $3$ and the sum of the squares of its terms is also $3$,what are the first term and the common ratio of the series?

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