Sum to infinity of progression 9 - 3 + 1 - frac{1}{3} + ..... is

English

Gujarati

Hindi

8. Sequences and Series

easy

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

A

$9$

B

$9/2$

C

$27/4$

D

$15/2$

Solution

(c) Infinite series $9 – 3 + 1 – \frac{1}{3}……\infty $ is a

$G.P.$ with $a = 9,r = \frac{{ – 1}}{3}$

${S_\infty } = \frac{a}{{1 – r}} = \frac{9}{{1 + \left( {\frac{1}{3}} \right)}} = \frac{{9 \times 3}}{4} = \frac{{27}}{4}$.

Standard 11

Mathematics

Share

Similar Questions

If the ${10^{th}}$ term of a geometric progression is $9$ and ${4^{th}}$ term is $4$, then its ${7^{th}}$ term is

easy

If $a,\;b,\;c$ are in $A.P.$, $b,\;c,\;d$ are in $G.P.$ and $c,\;d,\;e$ are in $H.P.$, then $a,\;c,\;e$ are in

hard

For a sequence $ < {a_n} > ,\;{a_1} = 2$ and $\frac{{{a_{n + 1}}}}{{{a_n}}} = \frac{1}{3}$. Then $\sum\limits_{r = 1}^{20} {{a_r}} $ is

easy

If the first and the $n^{\text {th }}$ term of a $G.P.$ are $a$ and $b$, respectively, and if $P$ is the product of $n$ terms, prove that $P^{2}=(a b)^{n}$

medium

If ${s_n} = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + …….. + \frac{1}{{{2^{n – 1}}}}$ , then the least integral value of $n$ such that $2 – {s_n} < \frac{1}{{100}}$ is

normal