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

Sum $1 + 2a + 3a^{2} + 4a^{3} + ....$ to $n$ terms.

  1. $\dfrac{1+(a^{n})}{(a-1)^{2}}-\dfrac{na^{n}}{1+a}$
  2. $\dfrac{1-2(a^{n})}{(a-1)^{2}}+\dfrac{na^{n}}{1-2a}$
  3. $\dfrac{1-(a^{n})}{(a-1)^{2}}-\dfrac{na^{n}}{1-a}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $S=1+2a+3a^{2}+4a^{3}+$   ..........$+na^{n-1}$
Multiply both sides by $a$, we get
$Sa=0+a+2a^{2}+3a^{3}$   .............$(n-1)a^{n-1}+na^{n}$
Subtract both equations,
$S(1-a)=1+a+a^{2}+a^{3}$    .............$a^{n-1}-na^{n}$
Clearly above series is G.P
Common ratio $= a$
$S(1-a)=\dfrac{1(a^{n}-1)}{a-1}-na^{n}$
$S=\dfrac{1-(a^{n})}{(a-1)^{2}}-\dfrac{na^{n}}{1-a}$

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

The sum to infinity of the terms of an infinite geometric progression is $6$. The sum of the first two terms is $4\dfrac {1}{2}$. The first term of the progression is

  1. $3$ or $1\dfrac {1}{2}$
  2. $1$
  3. $2\dfrac {1}{2}$
  4. $6$
  5. $9$ or $3$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Let the first two terms be $a$ and $ar$ where $-1 < r < 1$.
$\therefore a(1 + r) = 4\dfrac {1}{2}$
Since $s = a/(1 - r) = 6, a = 6(1 - r)$
$\therefore 6(1 - r)(1 + r) = 4\dfrac {1}{2}; \therefore r = \pm \dfrac {1}{2}; \therefore a = 3$ or $9$.

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

The sum of $2n$ terms of a series of which every even term is $'a'$ times the terms before it, and every odd term $'c'$ times the terms before it, the first term being unity, is

  1. $\dfrac { \left( 1-a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac-1 }$
  2. $\dfrac { \left( 1+a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac+1 }$
  3. $\dfrac { \left( 1+a \right) \left( { a }^{ n }{ c }^{ n }-1 \right) }{ ac-1 }$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} { T _{ 1 } }=1 \ { T _{ 2 } }=a \ { T _{ 3 } }=Ca \ { T _{ 4 } }=C{ a^{ 2 } } \ { T _{ 2n } }=a\frac { { 2n } }{ 2 } \cdot C\frac { { 2n } }{ 2 } -1={ a^{ n } }{ C^{ n-1 } } \ { 5 _{ 2n } }=1+a+ca.....{ a^{ n } }{ C^{ n-1 } } \ =1+\left[ { a+c{ a^{ 2 } }+{ c^{ 2 } }{ a^{ 3 } }...{ a^{ n } }{ c^{ n-1 } } } \right]  \ +\left[ { ca+{ c^{ 2 } }{ a^{ 2 } }+{ c^{ 3 } }{ a^{ 3 } }.....{ c^{ n-1 } }{ a^{ n-1 } } } \right]  \ =1+\frac { { a\left( { { a^{ n } }{ c^{ n-1 } } } \right)  } }{ { ac-1 } } +\frac { { ac\left( { { a^{ n-1 } }{ c^{ n-1 } }-1 } \right)  } }{ { ac-1 } }  \ =\frac { { \left( { { a^{ n } }{ c^{ n } }-1 } \right) \left( { a+1 } \right)  } }{ { ac-1 } }  \end{array}$

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

The sum of $10$ terms of the series $0.7 + .77 + .777 + \ldots \ldots \ldots$ is

  1. $\dfrac { 7 } { 9 } \left( 89 + \dfrac { 1 } { 10 ^ { 10 } } \right)$
  2. $\dfrac { 7 } { 81 } \left( 89 + \dfrac { 1 } { 10 ^ { 10 } } \right)$
  3. $\dfrac { 7 } { 81 } \left( 89 + \dfrac { 1 } { 10 ^ { 9 } } \right)$
  4. $\dfrac { 7 } { 9 } \left( 89 + \dfrac { 1 } { 10 ^ { 9 } } \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$0.7+0.77+0.777+......$
$=7\left( 0.1+0.11+0.111+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 0.9+0.99+0.999+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 1-0.1+1-0.1+1-0.001+...... \right) $
$=\dfrac { 7 }{ 9 } \left( 10-\left( 0.1+0.01+0.001+...... \right)  \right) $
$=\dfrac { 7 }{ 9 } \left( 10-\dfrac { 0.1\left( 1-{ 10 }^{ -10 } \right)  }{ 1-0.1 }  \right) =\frac { 7 }{ 9 } \left( 10-\dfrac { 0.1\left( { 10 }^{ 10 }-1 \right)  }{ 0.9\times { 10 }^{ 10 } }  \right) =\dfrac { 7 }{ 9 } \left( 10-\dfrac { 1 }{ 9 } +\dfrac { 1 }{ 9\times { 10 }^{ 10 } }  \right) $
$=\dfrac { 7 }{ 9 } \left( \dfrac { 89 }{ 9 } +\dfrac { 1 }{ 9\times { 10 }^{ 10 } }  \right) $
$=\dfrac { 7 }{ 81 } \left( 89+\dfrac { 1 }{ { 10 }^{ 10 } }  \right) $      [B]
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The sum of series $\displaystyle \frac{3}{4} + \frac{15}{16} + \frac{63}{64}+ ..... $ up to $n$ terms is

  1. $\displaystyle n - \frac{4^n}{3} - \frac{1}{3}$
  2. $\displaystyle n + \frac{4^{-n}}{3} - \frac{1}{3}$
  3. $\displaystyle n + \frac{4^n}{3} - \frac{1}{3}$
  4. $\displaystyle n - \frac{4^{-n}}{3} - \frac{1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For $n=1$, we have
$\displaystyle n - \dfrac{4^n }{3} - \dfrac{1}{3} = 1  - \dfrac{4}{3} - \dfrac{1}{3} = - \dfrac{2}{3}$
$\displaystyle n + \dfrac{4^n}{3} - \dfrac{1}{3} = 1 + \dfrac{4}{3} - \dfrac{1}{3} = 2$
$n - \displaystyle \dfrac{4^{-n}}{3} + \dfrac{1}{3} = 1 - \dfrac{4^{-1}}{3} + \dfrac{1}{3}= \dfrac{5}{4}$
Also, for $n = 2$, we have
$ \displaystyle n + \dfrac{4^{-n}}{3} - \dfrac{1}{3} = 2 + \dfrac{1}{48} - \dfrac{1}{3} = \dfrac{27}{16}$ and $\displaystyle \dfrac{3}{4} + \dfrac{15}{16} = \dfrac{27}{16}$
Hence, option (b) is correct.
ALTER We have,
$\displaystyle \dfrac{3}{4} + \dfrac{15}{16} + \dfrac{63}{64}+ ..... $ to n terms
$= \displaystyle \dfrac{2^2 - 1}{2^2} + \dfrac{2^4 - 1}{2^4} + \dfrac{2^6 - 1}{2^6}+ .... $ to n terms.
$= \displaystyle \left ( 1 - \dfrac{1}{2^2} \right ) + \left ( 1 - \dfrac{1}{2^4} \right ) + \left( 1 - \dfrac{1}{2^6} \right ) + ..... $ to n terms
$= n - \left \{ \dfrac{1}{2^2} + \dfrac{1}{2^4} + \dfrac{1}{2^6} + .... \text{to n terms} \right \}$
$= n \displaystyle - \dfrac{1}{2^2} \left \{ \dfrac{1 - \left (\dfrac{1}{2^2} \right )^n }{1 - \dfrac{1}{2^2}} \right \}$
$= \displaystyle n - \dfrac{1}{3} (1 - 4^{-n})$
$= n + \displaystyle \dfrac{4^{-n}}{3} - \dfrac{1}{3}$

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

If the sum of $n$ terms of a GP (with common ratio $r$) beginning with the $\displaystyle p^{th}$ term is $k$ times the sum of an equal number of the same series beginning with the $\displaystyle q^{th}$ term, then the value of $k$ is

  1. $\displaystyle r^{p/q}$
  2. $\displaystyle r^{q/p}$
  3. $\displaystyle r^{p-q}$
  4. $\displaystyle r^{p+q}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$p^{th}$ term of the series  $=ar^{p-1}$     ($a$ is first term)

$q^{th }$ term of th series $= ar^{q-1}$
Sum of $n$ term beginning with $p^{th} $ term 
$=\dfrac{ar^{p-1}(r^n - 1)}{r-1}$
Sum of $n$ term beginning with $q^{th} $ term 
$=\dfrac{ar^{q-1}(r^n - 1)}{r-1}$
Sum of $n$ term beginning with $p^{th} $ term $= k$ (sum of $n$ term beginning with $q^{th} $ term )
Thus $\dfrac{ar^{p-1}(r^n - 1)}{r-1}$$=k\dfrac{ar^{q-1}(r^n - 1)}{r-1}$
$\Rightarrow k = \dfrac{r^{p-1}}{r^{q-1}}$
$\Rightarrow k= r^{p-q}$

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

In a $G.P$. the ratio of the sum of the first eleven terms to the sum of last eleven terms is $\displaystyle \frac{1}{8}$ and the ratio of the sum of all terms without the first nine to the sum of all the terms without the last nine is $2$. Then the number of terms of the $G.P$ is

  1. $15$
  2. $43$
  3. $38$
  4. $56$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have: 
$\dfrac { \frac { a({ r }^{ 11 }-1) }{ r-1 }  }{ \frac { a{ r }^{ n-11 }({ r }^{ 11 }-1) }{ r-1 }  } =\dfrac { 1 }{ 8 }$
$\Rightarrow { r }^{ n-11 }=8$ ...(i)
Also:
$\dfrac { \frac { a{ r }^{ 9 }({ r }^{ 11 }-1) }{ (r-1) }  }{ \frac { a({ r }^{ n-9 }-1) }{ (r-1) }  } =2$
$\Rightarrow { r }^{ 9 }=2 $
$\Rightarrow r={ 2 }^{ \frac { 1 }{ 9 }  }$ ...(ii)
Substituting (ii) in (i):
${ 2 }^{ \frac { n-11 }{ 9 }  }={ 2 }^{ 3 }$
$\Rightarrow \dfrac { n-11 }{ 9 } =3$
$\Rightarrow n=38$
Hence, (c) is correct.

Multiple choice maths average arithmetic mean of ap introduction to averages means

Mean of the first $n$ terms of the A.P. $a, (a + d), (a + 2d), ........$ is

  1. $\displaystyle a + \frac{nd}{2}$
  2. $\displaystyle a + \frac{(n - 1)d}{2}$
  3. $a + (n - 1) d$
  4. $a + nd$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Required mean $= \displaystyle \frac{a + (a + d) + (a + 2d) + ....... + { a + (n - 1) d }}{n}$
$\displaystyle = \frac{\displaystyle \frac{n}{2} [a + a + (n - 1) d]}{n} = a + \frac{(n - 1)d}{2}$

Multiple choice maths average arithmetic mean of ap introduction to averages means

If $n^{th}$ term of AP is $4n+1$, then AM of $11^{th}$ to $ 20^{ th}$ terms is 

  1. $61.5$
  2. $63$
  3. $63.5$
  4. $62$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given: $t _{n}=4n+1$

$A.M.=\cfrac{t _{11}+t _{20}}{2}$
           $=\cfrac{4\times 11+1+4\times 20+1}{2}$
           $=63$

Multiple choice maths average arithmetic mean of ap introduction to averages means

If  $n^{th}$ term of AP is $t _n=4n+1$. Find mean of first $10$ terms.   

  1. $85$
  2. $95$
  3. $23$
  4. $7.5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given:
$t _{n}=4n+1$
$\therefore A.M.=\dfrac{\sum _{n=1}^{10}4n+1}{n}$
               $=\dfrac{2n(n+1)+n}{n}$
               $=2n+3$
               $=2\times 10+3$
               $=23$
Multiple choice maths average arithmetic mean of ap introduction to averages means

Sum of $4$ numbers in GP is $60$. And the AM of first and last no. is $18$ find the first term and common difference of the GP

  1. $a=4, r=2$
  2. $a=32, r=\dfrac {1}{2}$
  3. $a=3, r=1$
  4. $a=6, r=3$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

Let 4 numbers in G.P. be $a,ar,{ ar }^{ 2 }{ ,ar }^{ 3 }\ a(1+r+{ r }^{ 2 }+{ r }^{ 3 })=60\ \cfrac { a+{ ar }^{ 3 } }{ 2 } =18=>a+{ ar }^{ 3 }=36\ a(1+r)(1+{ r }^{ 2 })=60\ a(1+r)(1+{ r }^{ 2 }-r)=36\ \cfrac { 1+{ r }^{ 2 }-r }{ 1+{ r }^{ 2 } } =\cfrac { 36 }{ 60 } \ 5+5{ r }^{ 2 }-5r=3+3{ r }^{ 2 }\ 2{ r }^{ 2 }-5r+2=0\ r=2,\cfrac { 1 }{ 2 } \ if\quad r=2,a(1+8)=36=>a=4\ if\quad r=\cfrac { 1 }{ 2 } ,a(1+\cfrac { 1 }{ 8 } )=36=>a=32\ a=4,r=2\quad (or)\quad a=32,r=\cfrac { 1 }{ 2 } $

Multiple choice maths average arithmetic mean of ap introduction to averages means

Find the arithmetic mean of the series $1, 3, 5,...........
(2n - 1)$

  1. $n$
  2. $2n$
  3. $n/2$
  4. $n - 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the given Arithmetic Progression,
First term $ = a  = 1 $
Common difference $ = 3 - 1 = 2 $

Let $ 2n-1 $ be the $  k $ th term.

Then $ {x} _{n} = a + (n-1)d $
$ => 2n-1 = 1 + (k-1)2  $
$ => 2n -1 = 1 +2k-2 $
$ => k = n $

So,  $ 2n-1 $ is the $ n $ th term.

Now, Sum of the given series upto 'n' terms $ = \frac {n}{2} (2a+(n-1)d) =n^2 $

so mean is = sum of series/total no of terms
$= n^2/n=n$

Multiple choice maths average arithmetic mean of ap introduction to averages means

Find the arithmetic mean of the series: $1,3,5 ........... (2n - 1)$

  1. $n$
  2. $2n$
  3. $\dfrac n2$
  4. $n - 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$1,3,5,..., (2n-1)$

Total number of terms: $n$
Arithmetic mean: $\dfrac{1+3+5+\cdots+(2n-1)}{n}$

Using sum of first n terms of an AP:
$1+3+5+\cdots+(2n-1) = n^2$
$\therefore$ AM = $\dfrac{n^2}n = n$

Multiple choice maths average arithmetic mean of ap introduction to averages means

What is the average of the first $300$ terms of the given sequence?
$1, -2, 3, -4, 5, -6, ....., n.(-1)^{n + 1}$

  1. $-1$
  2. $0.5$
  3. $0$
  4. $-0.5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Avg $=\cfrac{[1+3+5+7+...+(2n-1)]-[2+4+6+...+2n]}{n}$
Here $n=300$
No. of even terms $=150$
No. of odd terms $=150$
Now,
$1+3+5+7+...+150\;terms \\ S _{n}=\cfrac{n}{2}[2a+(n-1)d] \\ S _{n1}=\cfrac{150}{2} [2\times 1 +149\times 2]=22500$
and $2+4+6+8+....+150\;terms \\ S _{n}=\cfrac{n}{2}[2a+(n-1)d] \\ S _{n2}=\cfrac{150}{2}[2\times 2+149\times 2]=22650$
Average $=\cfrac{S _{n1}-S _{n2}}{300}=\cfrac{22500-22650}{300} \\ =-0.5$