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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

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

Vedclass · Answer

(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

Similar Questions

Consider two positive numbers $a$ and $b$. If the arithmetic mean of $a$ and $b$ exceeds their geometric mean by $\frac{3}{2}$ and the geometric mean of $a$ and $b$ exceeds their harmonic mean by $\frac{6}{5}$,then the absolute value of $(a^2 - b^2)$ is equal to

The sum of the series $1^2 \cdot 2 + 2^2 \cdot 3 + 3^2 \cdot 4 + \dots$ to $n$ terms is:

If the sum of first $n$ terms of an $A.P.$ is $cn(n - 1)$,where $c \neq 0$,then the sum of the squares of these terms is

The sequence $\frac{5}{\sqrt{7}}, \frac{6}{\sqrt{7}}, \sqrt{7}, \dots$ is

Let $\alpha$ and $\beta$ be the roots of $x^{2}-3x+p=0$ and $\gamma$ and $\delta$ be the roots of $x^{2}-6x+q=0$. If $\alpha, \beta, \gamma, \delta$ form a geometric progression,then the ratio $(2q+p):(2q-p)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module