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 $9 - 3 + 1 - \frac{1}{3} + .....$ is a Geometric Progression $(G.P.)$.Here,the first term $a = 9$.The common ratio $r = \frac{-3}

Vedclass · Accepted Answer

$27/4$

Vedclass · Answer

(C) The given series $9 - 3 + 1 - \frac{1}{3} + .....$ is a Geometric Progression $(G.P.)$.Here,the first term $a = 9$.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 $9 - 3 + 1 - \frac{1}{3} + .....$ is

Similar Questions

If $\log _{5} 2, \log _{5}(2^{x}-3)$ and $\log _{5}(\frac{17}{2}+2^{x-1})$ are in $A.P.$,then the value of $x$ is:

If $\sum\limits_{r = 0}^{25} {\left( {^{50}{C_r} \cdot ^{50 - r}{C_{25 - r}}} \right) = K\left( {^{50}{C_{25}}} \right)}$,then $K$ is equal to

If the sum of the first $n$ terms of the series $\sqrt{3} + \sqrt{75} + \sqrt{243} + \sqrt{507} + \dots$ is $435\sqrt{3}$,then $n$ equals

$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^{n - 1} {\frac{k}{{{2^{n + k}}}}} } $ is equal to

If $S_n = \frac{n(n + 1)(n + 2)}{6}$,then $\sum_{n = 1}^\infty \frac{1}{t_n} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module