The sum of infinite terms of the following se | vedclass

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

Question

(D) Let the sum to infinity of the arithmetico-geometric series be $S = 1 + 4 \cdot \frac{1}{5} + 7 \cdot \frac{1}{5^2} + 10 \cdot \frac{1}{5^3} + \do

Vedclass · Accepted Answer

$\frac{35}{16}$

Vedclass · Answer

(D) Let the sum to infinity of the arithmetico-geometric series be $S = 1 + 4 \cdot \frac{1}{5} + 7 \cdot \frac{1}{5^2} + 10 \cdot \frac{1}{5^3} + \dots$$\frac{1}{5}S = \frac{1}{5} + 4 \cdot \frac{1}{5^2} + 7 \cdot \frac{1}{5^3} + \dots$Subtracting the two equations:$(1 - \frac{1}{5})S = 1 + 3 \cdot \frac{1}{5} + 3 \cdot \frac{1}{5^2} + 3 \cdot \frac{1}{5^3} + \dots$$\frac{4}{5}S = 1 + 3 \left( \frac{1}{5} + \frac{1}{5^2} + \dots ight)$Using the sum of an infinite geometric series formula $S_{\infty} = \frac{a}{1-r}$ where $a = \frac{1}{5}$ and $r = \frac{1}{5}$:$\frac{4}{5}S = 1 + 3 \left( \frac{1/5}{1 - 1/5} ight) = 1 + 3 \left( \frac{1/5}{4/5} ight) = 1 + 3 \left( \frac{1}{4} ight) = 1 + \frac{3}{4} = \frac{7}{4}$$S = \frac{7}{4} 	imes \frac{5}{4} = \frac{35}{16}$.

The sum of infinite terms of the following series $1 + \frac{4}{5} + \frac{7}{5^2} + \frac{10}{5^3} + \dots$ will be

Similar Questions

If $\tan \theta = \sqrt{\frac{3}{2}}$,the sum of the infinite series $1 + 2(1 - \cos \theta) + 3(1 - \cos \theta)^2 + 4(1 - \cos \theta)^3 + \dots \infty$ is

If $S(x) = (1+x) + 2(1+x)^2 + 3(1+x)^3 + \ldots + 60(1+x)^{60}$,$x \neq 0$,and $(60)^2 S(60) = a(b)^b + b$ where $a, b \in N$,then $(a+b)$ is equal to:

$1 + \frac{3}{2} + \frac{5}{2^2} + \frac{7}{2^3} + \dots \infty$ is equal to

If the $r^{th}$ term of a series is $(2r + 1)2^{-r}$,what is the sum of its infinite terms?

The sum $\sum\limits_{k = 1}^{20} {k\frac{1}{{{2^k}}}} $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module