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

The sum of $2n$ terms of a geometric progression whose first term is $'a'$ and common ratio $'r'$ is equal to the sum of $n$ terms of a geometric progression whose first term is $'b'$ and common '$r^{2}$'. then $b$ is equal to

  1. The sum of the first two terms of the first series.

  2. The sum of the first and last terms of the first series.

  3. The sum of the last two terms of the first series.

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that

$\begin{array}{l} \dfrac { { a\left( { { r^{ 2n } }-1 } \right)  } }{ { r-1 } } =\dfrac { { b{ { \left( { { r^{ 2 } } } \right)  }^{ n } }-1 } }{ { { r^{ 2 } }-1 } }  \ \Rightarrow \dfrac { { a\left( { { r^{ 2n } }-1 } \right)  } }{ { r-1 } } =\dfrac { { b\left( { { r^{ 2n } }-1 } \right)  } }{ { (r-1)\left( { r+1 } \right)  } }  \ \Rightarrow b=a\left( { r+1 } \right)  \ \Rightarrow b=a+ar \end{array}$
$b$= sum of first two term of the first series.

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

In a infinite G.P. , the sum of first three terms is 70. If the extreme terms are multiplied by 4 and the middle term is multiplied by 5, the resulting terms form an A.p. then the sum to infinite terms of G.p.   

  1. 120

  2. -40

  3. 160

  4. 80
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,

The three numbers be $a,ar\,and\,a{{r}^{2}}$.

Given that,


$ a+ar+a{{r}^{2}}=70 $

$ \Rightarrow a\left( 1+r+{{r}^{2}} \right)=70\,\,......\,\,\left( 1 \right) $


Also given that,

$4a,\,5ar\,and\,4a{{r}^{2}}$ in an A.P.

$\begin{align}

$ \Rightarrow 2\left( 5ar \right)=4a+4a{{r}^{2}} $

$ \Rightarrow 5r=2+2{{r}^{2}} $

$ \Rightarrow 2{{r}^{2}}-5r+2=0 $

$ \Rightarrow 2{{r}^{2}}-\left( 4+1 \right)r+2=0 $

$ \Rightarrow 2{{r}^{2}}-4r-1r+2=0 $

$ \Rightarrow 2r\left( r-2 \right)-1\left( r-2 \right)=0 $

$ \Rightarrow \left( r-2 \right)\left( 2r-1 \right)=0 $

$ \Rightarrow r-2=0,\,\,2r-1=0 $

$ \Rightarrow r=2,\,\,r=\dfrac{1}{2} $

From (1) we get,

$a=10\,\,\,at\,\,\,r=2$

And $a=40\,\,\,at\,\,\,r=\dfrac{1}{2}$

Sum of this series

$ =\dfrac{a}{1-r} $

$ =\dfrac{40}{1-2} $

$ =-40 $

This is the answer.

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

Sum of the first five terms of the geometric series $1 + \dfrac {2}{3} + \dfrac {4}{9} + $....is 

  1. $\dfrac {211}{81}$
  2. $\dfrac {81}{211}$
  3. $-\dfrac {211}{81}$
  4. $-\dfrac {81}{211}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle { s } _{ 5 }=\frac { 1\times \left[ { 1-\left( { 2 }/{ 3 } \right)  }^{ 5 } \right]  }{ 1-\left( { 2 }/{ 3 } \right)  } =\frac { 211 }{ 81 } $

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

The sum of the geometric sequence is given as $S=\cfrac{a(1-r^n)}{1-r}$, where $r$ is the

  1. constant

  2. term

  3. common difference

  4. common ratio
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know $S=\dfrac {a(1-r^n)}{1-r}$
In a geometric sequence, the ratio of terms is constant and is known as the common ratio $(r)$.
So, in the formula for summation, $r$ is common ratio.
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

$x, 2x, 4x, . . .$
The first term in the sequence above is $x$, and each term thereafter is equal to twice the previous term. Find the sum of the first five terms of this sequence.

  1. $10x$
  2. $15x$
  3. $30x$
  4. $31x$
  5. $32x$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The first term in given Geometric series, $a _1=x$
The common ratio $r$ $=\dfrac{4x}{2x}=2$
No. of terms, $n$ $=5$
Applying sum of GP formula,
$S _n=\dfrac{a(1-r^n)}{1-r}$
      $=\dfrac{x(1-2^5)}{1-2}$
      $=\dfrac{x(1-32)}{-1}=31x$
Hence,option D is correct.

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 following G.P. to $n$ terms $0.5 + 0.55 + 0.555 + 0.5555 + .....$

  1. $\dfrac {5}{9}\left[9n-1+\dfrac {1}{10^n}\right]$
  2. $\dfrac {5}{81}\left[5n-1-\dfrac {1}{10^n}\right]$
  3. $\dfrac {5}{81}\left[9n-1+\dfrac {1}{10^n}\right]$
  4. $-\dfrac {5}{9}\left[9n-1+\dfrac {1}{10^n}\right]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The series is 0.5 + 0.55 + 0.555... = (5/9) * (0.9 + 0.99 + 0.999...) = (5/9) * [(1-0.1) + (1-0.01) + (1-0.001)...]. This simplifies to (5/9) * [n - (0.1 + 0.01 + ... + 0.1^n)]. The sum of the geometric part is (1/10)(1-(1/10)^n)/(1-1/10) = (1/9)(1-1/10^n). Multiplying through leads to the correct expression.

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

The sum of the first three terms of an increasing G.P. is $13$ and their product is $27$. The sum of the first $5$ terms is,

  1. $323$
  2. $363$
  3. $109$
  4. $254$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the G.P be $\dfrac{a}{r}, a, ar$.


So,
$\dfrac{a}{r}+ a+ ar=13$

And
$\dfrac{a}{r}\times  a\times  ar=27$

$a^3=27$
$a=3$

Therefore,
$\dfrac{3}{r}+ 3+ 3r=13$

$3+ 3r+ 3r^2=13r$

$3r^2-10r+3=0$

$3r^2-9r-r+3=0$

$3r(r-3)-1(r-3)=0$

$(3r-1)(r-3)=0$

$r=\dfrac{1}{3}, 3$

So, $r=3$

So, the G.P is $1, 3, 9$.

Now, the sum of first five terms
$=\dfrac{3(3^5-1)}{3-1}$

$=\dfrac{3(243-1)}{2}$

$=\dfrac{3(242)}{2}$

$=3(121)=363$

Hence, this is the answer.

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

The sum of sequence $0.15,0.015,0.0015,.....$ upto 20 term is ?

  1. $\dfrac{1}{6}[1-(0.1)^{20}]$
  2. $\dfrac{1}{6}[1+(0.1)^{20}]$
  3. $\dfrac{1}{3}[1-(0.1)^{20}]$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the sequence in Geometric progression 

where, $a=0.15;\ r=\dfrac { 0.015 }{ 0.15 } =0.1;\ n=20\ \therefore { S } _{ n }=\dfrac { a({ r }^{ n }-1) }{ (r-1) } \ =\dfrac { 0.15({ \left( 0.1 \right)  }^{ 20 }-1) }{ 0.1-1 } \ =\dfrac { 1 }{ 6 } \left( 1-{ (0.1) }^{ 20 } \right) $

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 GP $\frac { 1 } { 2 } + \frac { 1 } { 4 } + \frac { 1 } { 8 } + \ldots$ is-

  1. $\frac { 2 ^ { 10 } - 1 } { 2 ^ { 10 } }$
  2. $\frac { 2 ^ { 9 } - 1 } { 2 ^ { 9 } }$
  3. $\frac { 2 ^ { 10 } - 1 } { 2 ^ { 9 } }$
  4. $\frac { 2 ^ { 9 } - 1 } { 2 ^ { 10 } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...$ is in G.P
where $a=\dfrac{1}{2}$ and $r=\dfrac{\dfrac{1}{4}}{\dfrac{1}{2}}=\dfrac{1}{2}$
${S} _{n}=\dfrac{a\left({r}^{n}-1\right)}{r-1}$
${S} _{10}=\dfrac{\dfrac{1}{2}\left({\left(\dfrac{1}{2}\right)}^{10}-1\right)}{\dfrac{1}{2}-1}$
${S} _{10}=\dfrac{\dfrac{1}{2}\left({\left(\dfrac{1}{2}\right)}^{10}-1\right)}{\dfrac{-1}{2}}$

$=1-\dfrac{1}{{2}^{10}}$
$=\dfrac{{2}^{10}-1}{{2}^{10}}$


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

The geometric series $a+ar+ar^{2}+ar^{3}+......\infty$ has sum $7$ and the terms involving odd powders of $r$ has sum $'3'$, then the value of $(a^{2}-r^{2})$ is-

  1. $\dfrac{5}{4}$
  2. $\dfrac{5}{2}$
  3. $\dfrac{25}{4}$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Sum of geometric series = $\dfrac{a(r^{n}-1)}{r-1}=7$
Also odd powers of r terms :-
$ar,ar^{3},ar^{5},ar^{7}$
$\therefore sum=\dfrac{ar(r^{2n}-1)}{r^{2}-1}$
$\therefore $ sum of infinite G.P = $\dfrac{a}{1-r}=7$ _________(1)
$\therefore $ sum of infinite second = $\dfrac{ar}{1-r^{2}}=3$ _____(2)
$\therefore $ From (1) & (2)
$\dfrac{1+r}{r}=\dfrac{7}{3} ; r=\dfrac{3}{4}$
$\therefore a=\dfrac{7}{4}$
$\therefore a^{2}-r^{2}=\dfrac{5}{2}$
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 $10$ terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+...$ is

  1. $121(\sqrt{6}+\sqrt{2})$
  2. $243(\sqrt{3}+1)$
  3. $\cfrac{121}{\sqrt{3}-1}$
  4. $242(\sqrt{3}-1)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$s=\sqrt{2}+\sqrt{6}+\sqrt{18}+....$

$\Rightarrow s=\sqrt{2}+\sqrt{2}\times \sqrt{3}+\sqrt{2}(\sqrt{3})^2+.....$

This is a G.P with $1$st term $(a)=\sqrt{2}$ and common ratio$(r)=\sqrt{3}$

Sum of $10$ terms of this G.P., $S=\dfrac{a(r^{10}-1)}{r-1}$

$=\dfrac{\sqrt{2}((\sqrt{3})^{10}-1)}{\sqrt{3}-1}$

$=\dfrac{\sqrt{2}(242)}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}$

$=121\times \sqrt{2}(\sqrt{3}+1)$

$=121(\sqrt{6}+\sqrt{2})$

$\Rightarrow (A)$ Option. 

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

If ${S} _{n}=\sum _{ r=1 }^{ n }{ \cfrac { 1+2+{ 2 }^{ 2 }+..Sum\quad to\quad r\quad terms }{ { 2 }^{ r } }  } $, then ${S} _{n}$ is equal to 

  1. ${2}^{n}-n-1$
  2. $1-\cfrac{1}{{2}^{n}}$
  3. $n-1+\cfrac{1}{{2}^{n}}$
  4. ${2}^{n}-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\displaystyle  \rightarrow S _{n} = \sum _{r=1}^{n}\left(\dfrac{\frac{2^{r}-1}{2-1}}{2^{r}}\right)$ sum of G.P

$\displaystyle  \Rightarrow S _{n} = \sum _{r=1}^{n}(1-2^{-r})$

$ \displaystyle \Rightarrow S _{n} = n- \sum _{r=1}^{n}2^{-r}$

$ \displaystyle \Rightarrow S _{n} = n -\left(2^{-1}(\frac{2^{-n}-1}{2^{-1}-1})\right)$

$\displaystyle  \Rightarrow S _{n} = n-\left(\frac{1}{2}(\frac{1-2^{n}}{2^{n}(\frac{-1}{2})})\right)$

$\displaystyle  \Rightarrow S _{n} = n+\left(\frac{1-2^{n}}{2^{n}}\right) = n-1+\frac{1}{2^{n}}$

$ \Rightarrow (C)$