The sum of an infinite geometric series is 3. | vedclass

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Question

(A) Let the first term of the infinite geometric series be $a$ and the common ratio be $r$. The sum of an infinite geometric series is given by $S = \

Vedclass · Accepted Answer

$\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, .....$

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(A) Let the first term of the infinite geometric series be $a$ and the common ratio be $r$. The sum of an infinite geometric series is given by $S = \frac{a}{1-r}$.Given,$\frac{a}{1-r} = 3$,so $a = 3(1-r)$ ..... $(i)$.The series formed by the squares of its terms is $a^2, a^2r^2, a^2r^4, .....$. This is also an infinite geometric series with first term $a^2$ and common ratio $r^2$.The sum of this series is $\frac{a^2}{1-r^2} = 3$ ..... $(ii)$.Substituting $a = 3(1-r)$ into $(ii)$:$\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}$.Substituting $r = \frac{1}{2}$ into $(i)$:$a = 3(1 - \frac{1}{2}) = 3(\frac{1}{2}) = \frac{3}{2}$.The series is $a, ar, ar^2, .....$,which is $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, .....$.

The sum of an infinite geometric series is $3$. $A$ series,which is formed by squares of its terms,has the sum also $3$. The first series is

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