in Quantitative Aptitude edited by
115 views
1 vote
1 vote

An infinite geometric progression $a_{1},a_{2},a_{3},\dots\dots$ has the property that $a_n =3(a_{n+1}+a_{n+2}+\dots\dots)$ for every $n\geq 1$. If the sum $a_{1}+a_{2}+a_{3}+\dots\dots=32,$ then $a_{5}$ is 

  1. $1/32$
  2. $2/32$
  3. $3/32$
  4. $4/32$
in Quantitative Aptitude edited by
by
13.2k points 272 2046 2467
115 views

1 Answer

1 vote
1 vote

Given that,  

  • $a_1,a_2,a_3, \ldots$  in infinite geometric progression.
  •  $a_n=3(a_{n+1}+a_{n+2}+\ldots);\forall n\geq 1 \quad \longrightarrow (1)$

The sum   $a_1 + a_2 + a_3 + \ldots = 32$

$\Rightarrow \frac{a_1}{1-r} = 32 \quad \longrightarrow (2)$

From equation $(1)$, we have

$a_1 = 3(a_2+a_3+ \ldots)$

$\Rightarrow a_1=3\left(\frac{a_2}{1-r}\right)$

$\Rightarrow a_1 = 3 \left(\frac{a_1r}{1-r}\right)$

$\Rightarrow 1-r = 3r$

$\Rightarrow 4r = 1$

$\Rightarrow \boxed{r = \frac{1}{4}}$

Put the value of $r$ in equation $(2)$, we get.

$\Rightarrow \dfrac{a_1}{1-\frac{1}{4}} = 32$

$\Rightarrow \dfrac{a_1}{\frac{4-1}{4}} = 32$

$\Rightarrow \dfrac{4a_1}{3} = 32$

$\Rightarrow \boxed{a_1 = 24}$

We know that $n^{\text{th}}$ term of $\text{GP}$ series: $\boxed{a_n = ar^{n-1}},$ where $a$ and $r$ be the first term and common ratio respectively.

So, $a_5 = a_1r^{5-1}$

$\Rightarrow a_5 = 24\left(\frac{1}{4}\right)^{4}$

$\Rightarrow a_5 = 24\times\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}$

$\Rightarrow \boxed{a_5 = \frac{3}{32}}.$

Correct Answer $:\text{C}$


$\textbf{PS:}$

  • The sum of an infinite Geometric Progression whose first term $`a\text{'}$ and common ratio $`r\text{'}\; (-1 < r < 1 \;\text{i.e.,}\; |r| < 1)$ is $S_{\infty} = \dfrac{a}{1-r}.$
edited by
by
7.7k points 3 8 30
Answer:

Related questions

Ask
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true