Mathematics · Quantitative Aptitude

Sequences and Series

230 Questions

Sequences and series involve ordered lists of numbers and the sum of their terms. The questions primarily test knowledge of arithmetic progressions, geometric progressions, and infinite series. It is an essential part of quantitative aptitude that requires strong pattern recognition skills.

Arithmetic progressionGeometric progressionInfinite seriesSum of termsNumber sequences

Sequences and Series Questions

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

If $S$ is the sum to infinity of a $G.P.$ whose first terms is $a$ then the sum of the first $n$ terms is 

  1. $S\left(1-\dfrac{a}{S}\right)^{n}$
  2. $S\left[1-\left(1-\dfrac{a}{S}\right)\right]^{n}$
  3. $a\left[1-\left(1-\dfrac{a}{S}\right)\right]^{n}$
  4. $S\left[1-\left(1-\dfrac{S}{a}\right)\right]^{n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an infinite G.P., S = a / (1-r), so r = 1 - a/S. The sum of the first n terms is S_n = a(1-r^n) / (1-r). Substituting r = 1 - a/S and 1-r = a/S, we get S_n = a(1 - (1-a/S)^n) / (a/S) = S(1 - (1-a/S)^n).

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Let $\displaystyle S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$ find the sum of first $20$ terms of the series

  1. $\displaystyle \frac{2^{20}-1}{2^{20}}$
  2. $\displaystyle \frac{2^{19}-1}{2^{19}}$
  3. $\displaystyle \frac{2^{20}-1}{2^{19}}$
  4. $\displaystyle \frac{2^{19}-1}{2^{20}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$S=1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 8 } .......$ first $20$ terms

$n=20$ and series is in $GP$ with common difference $=\cfrac { \cfrac { 1 }{ 2 }  }{ 1 } =\cfrac { \cfrac { 1 }{ 4 }  }{ \cfrac { 1 }{ 2 }  } =\cfrac { 1 }{ 2 } $
$ a=1\quad r=\cfrac { 1 }{ 2 } $
Sum$=\cfrac { a(1-{ r }^{ n }) }{ 1-r } $  when$\quad r<1$
$ =\cfrac { 1(1-{ (\cfrac { 1 }{ 2 } ) }^{ 20 }) }{ 1-\cfrac { 1 }{ 2 }  } \ =\cfrac { (1-\cfrac { 1 }{ { 2 }^{ 20 } } ) }{ \cfrac { 1 }{ 2 }  } \ =2(1-\cfrac { 1 }{ { 2 }^{ 20 } } )\ =\cfrac { 2({ 2 }^{ 20 }-1) }{ { 2 }^{ 20 } } \ =(\cfrac { { 2 }^{ 20 }-1 }{ { 2 }^{ 19 } } )$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The $n^{th}$ term of the sequence 

$\displaystyle\frac{1}{100}$, $\displaystyle\frac{1}{10000}$, $\displaystyle\frac{1}{1000000}$, $\dots\dots$ is

  1. $(1000)^n$
  2. $10^{2n}$
  3. $10^{-2n}$
  4. $10^{-n}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given series is a Geometric Progression, with first terms $ a = \dfrac {1}{100} $ and common ratio $ r = \dfrac {T _2}{T _1} = \dfrac {\dfrac {1}{10000}}{\dfrac {1}{100}} = \dfrac {1}{100} $

For a GP, the $ nth $ term is given by $ T _n = ar^{n-1} =\dfrac {1}{100}  \times (\dfrac {1}{100})^{n-1} =(\dfrac {1}{100})^{n} = 10^{-2n}$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Find $S _n$, the sum of the first $n$ terms, for the following geometric series. $a _1=120, a _5= 1, r=-2$.

  1. $20.66$
  2. $40.66$
  3. $80.66$
  4. $100.66$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, first term, $a = 120$, common ratio, $r = -2$ and $a _5=1$
We know $S _n=\dfrac{a _1-a _nr}{1-r}$
$S _n=\dfrac{120-(-2)}{1-(-2)}$
$S _n=\dfrac{120-(-2)}{1-(-2)}$
$S _n=\dfrac{122}{3}$
$S _n=40.66$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Find the sum of the first $6$ terms of the geometric series $80 - 20 + 5 +.....$

  1. $63.984$
  2. $32.451$
  3. $54.876$
  4. $25.458$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

First term, $a$ is $80$
Common ratio, $r =$ $\dfrac{-20}{80}=\dfrac{-1}{4}$
$S _n=\dfrac{a(1-r^2)}{1-r}$
$S _n=\dfrac{80(1-(\frac{-1}{4})^2)}{1-\frac{-1}{4}}$
$S _n = \dfrac{79.98}{1.25}$
$S _n = 63.98$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Determine the sum of the first 8 terms of the G.P. $1, 2, 4, 8...$

  1. $256$
  2. $255$
  3. $254$
  4. $253$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given series is $1,2,4,8,....$
To find the sum of the first $S _n$ terms of a geometric sequence using the formula.
From given question, we have
$a = 1, r = 2, n = 8$
Therefore, $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _8 = \dfrac{1(1-(2)^{8})}{1-2}$
$\Rightarrow S _8 = \dfrac{1(-255)}{-1}$
$\Rightarrow S _8 = 255$
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

Evaluate the sum of the first nine terms of the geometric sequence $5, 10, 20,...$

  1. $1555$
  2. $2555$
  3. $3555$
  4. $4555$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given sequence is $5,10,20,....$
To find the sum of the first $S _n$ terms of a geometric sequence using the formula
Here $a = 5, r = 2, n = 9$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _9 = \dfrac{5(1-2^{9})}{1-2}$
$\Rightarrow S _9 = \dfrac{-2555}{-1}$
$\Rightarrow S _9 = 2555$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The sum of $6^{th}$ term in the geometric series $4, 12, 36...$ is

  1. $1456$
  2. $2456$
  3. $3456$
  4. $4456$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given sequence is $4,12, 36$
To find the sum of the first $S _n$ terms of a geometric sequence using the formula
Here $a = 4, r = 3, n = 6$
We know $S _n = \dfrac{a _1(1-r^n)}{1-r}$
$\Rightarrow S _6 = \dfrac{4(1-3^{6})}{1-3}$
$\Rightarrow S _6 = \dfrac{-2912}{-2}$
$\Rightarrow S _6 = 1456$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The sum of first $n$ terms of an G.P. is

  1. $S _n = \cfrac{a _1(1-r^n)}{1-r}$
  2. $S _n = \cfrac{a _1(1+r^n)}{1-r}$
  3. $S _n = \cfrac{a _1(1-r^n)}{1+r}$
  4. $S _n = \cfrac{a _1(1-r^n)}{r-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A GP can be written as:

$a,ar,ar^2, ar^3..............,ar^{n-1}$
$\text{sum} = a+ar+ar^2+ar^3+.........+ar^{n-1}$
$\text{sum} = a(r^{n-1}+r^{n-2}+r^{n-3}+r^{n-4}+...........+r+1)$
We know that:
$\dfrac{x^n -1}{x-1} = x^{n-1}+x^{n-2}+x^{n-3}+.............+x+1$
Thus $\text{sum} = a\left (\dfrac{r^n -1}{r-1}\right)$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

What is the sum of first eight terms of the series $1-\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } -\cfrac { 1 }{ 8 } +.....$?

  1. $\cfrac { 89 }{ 128 } $
  2. $\cfrac { 57 }{ 384 } $
  3. $\cfrac { 85 }{ 128 } $

  4. None of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given series is a sum of terms of GP with common ratio $-\dfrac{1}{2}$
Sum of $n$ terms of GP with common ratio $r$ and first term $a$ is $\dfrac { a(1-r^{ n }) }{ 1-r } $
Putting $a=1,n=8$ and $r=-\dfrac { 1 }{ 2 } $ in above equation, we have 

Sum $=\dfrac { 1(1-(-\frac { 1 }{ 2 } )^{ 8 }) }{ 1-(-\frac { 1 }{ 2 } ) } =\dfrac { 1-\frac { 1 }{ 256 }  }{ \frac { 3 }{ 2 }  } =\dfrac { 255 }{ 128\times 3 } =\dfrac { 85}{128} $
Hence, option C is correct