The sum to infinity of the progression 9 - 3 | vedclass

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

Question

(C) The given series $9 - 3 + 1 - \frac{1}{3} + \dots$ is a geometric progression $(G.P.)$.Here,the first term $a = 9$ and the common ratio $r = \frac

Vedclass · Accepted Answer

$27/4$

Vedclass · Answer

(C) The given series $9 - 3 + 1 - \frac{1}{3} + \dots$ is a geometric progression $(G.P.)$.Here,the first term $a = 9$ and the common ratio $r = \frac{-3}{9} = -\frac{1}{3}$.The sum to infinity of a $G.P.$ is given by the formula $S_{\infty} = \frac{a}{1 - r}$,provided $|r| Substituting the values,we get $S_{\infty} = \frac{9}{1 - (-1/3)} = \frac{9}{1 + 1/3} = \frac{9}{4/3} = \frac{9 	imes 3}{4} = \frac{27}{4}$.

The sum to infinity of the progression $9 - 3 + 1 - \frac{1}{3} + \dots$ is

Similar Questions

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

Let $\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \ldots$ up to $10$ terms and $\beta = \sum_{n=1}^{10} n^4$. If $4\alpha - \beta = 55k + 40$,then $k$ is equal to . . . . . . .

$\sum\limits_{m = 1}^n {{m^2}}$ is equal to

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = \frac{n}{n+1}$.

$\sum_{n=1}^5 n(n^2+n+1) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module