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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the first term be $a$ and the common ratio be $r$. The sum of the infinite geometric series is given by $S = \frac{a}{1-r} = 3$,so $a = 3(1-r) \dots (i)$.The series formed by squaring its terms is $a^2, a^2r^2, a^2r^4, \dots$,which is a geometric series with first term $a^2$ and common ratio $r^2$. Its sum is $\frac{a^2}{1-r^2} = 3$.Substituting $a = 3(1-r)$ into the second equation: $\frac{9(1-r)^2}{1-r^2} = 3$.Since $1-r^2 = (1-r)(1+r)$,we have $\frac{9(1-r)^2}{(1-r)(1+r)} = 3$,which simplifies to $\frac{3(1-r)}{1+r} = 1$.Solving for $r$: $3 - 3r = 1 + r$,so $4r = 2$,which gives $r = \frac{1}{2}$.Substituting $r = \frac{1}{2}$ into $(i)$: $a = 3(1 - \frac{1}{2}) = \frac{3}{2}$.The series is $\frac{3}{2}, \frac{3}{2}(\frac{1}{2}), \frac{3}{2}(\frac{1}{2})^2, \dots$,which is $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \frac{3}{16}, \dots$.

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

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