Mathematics · Quantitative Aptitude

Sequences and Series

226 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 sum to infinite terms of a gp sequence, progression and series maths

The sum of the series formed by the sequence $3, \sqrt{3}, 1....... $ upto infinity is : 

  1. $\frac {3\sqrt{3}(\sqrt{3}+1)}{2}$
  2. $\frac {3\sqrt{3}(\sqrt{3} - 1)}{2}$
  3. $\frac {3(\sqrt{3}+1)}{2}$
  4. $\frac {3(\sqrt{3}-1)}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the given series

$\ \cfrac { \sqrt { 3 }  }{ 3 } =\cfrac { 1 }{ \sqrt { 3 }  } \$
So it is a GP with common ratio $\cfrac{1}{\sqrt {3}}$
Formula for sum of infinite terms of GP $ =\cfrac { a }{ 1-r }$,  where
$a$ is first term and r is the common ratio
$\Rightarrow  { S } _{ \infty  }=\cfrac { 3 }{ 1-\cfrac { 1 }{ \sqrt { 3 }  }  }  =\cfrac { 3\sqrt { 3 }  }{ \sqrt { 3 } -1 }$

$=\cfrac { 3\sqrt { 3 } (\sqrt { 3 } +1) }{ (\sqrt { 3 } -1)(\sqrt { 3 } +1) }$ ..... [On rationalizing]

$ =\cfrac { 3\sqrt { 3 } (\sqrt { 3 } +1) }{ 2 } $
Hence, A is correct.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If the sum of an infinitely decreasing G.P. is $3$, and the sum of the squares of its terms is $\dfrac {9}{2}$, then the sum of the cubes of the terms is

  1. $\dfrac {105}{13}$
  2. $\dfrac {108}{13}$
  3. $\dfrac {729}{8}$
  4. $\dfrac {108}{9}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the GP be $a,ar,ar^2,ar^3,...$
The first term be $a$ and the common ratio be $r$. 
Then, it is given that:
$\text{Sum}=\dfrac{a}{1-r} = 3$    ....(1)
Sequence of squares of terms is $a^2, a^2r^2, a^2r^4,...$
$\text{Sum}=\dfrac{a^2}{1-r^2} = \dfrac{9}{2}$    ....(2)
Thus, we have:
$\dfrac{a^2\div a}{(1-r^2)\div (1-r)} = \dfrac{9 \div 3}{2} $
$\Rightarrow \dfrac{a}{1+r} = \dfrac{3}{2}$    ....(3)

Now divide equations (1) and (3),
$\dfrac{1-r}{1+r} = \dfrac{1}{2}$
$ \Rightarrow r = \dfrac{1}{3}$
Substituting this value of $r$ in equation (1), we get $a = 2$
Therefore, $\dfrac{a^3}{1-r^3} = \dfrac{8}{\left (1-\dfrac{1}{27}\right)} = \dfrac{108}{13}$
Thus, the answer is option B.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

Sum of the series ${9^{{1 \over 3}}} \times {9^{{1 \over 9}}} \times {9^{{1 \over {27}}}} \times .......$  is equal to

  1. $3$
  2. $9$
  3. $27$
  4. $81$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ 9 }^{ \cfrac { 1 }{ 3 }  }\times { 9 }^{ \cfrac { 1 }{ 9 }  }\times { 9 }^{ \cfrac { 1 }{ 27 }  }\times ...\infty $

$={ 9 }^{ \cfrac { 1 }{ 3 }  +\cfrac { 1 }{ 9 } +\cfrac { 1 }{ 27 } + ...}$
Let $S=\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 9 } +\cfrac { 1 }{ 27 } +...$
$=\cfrac { \cfrac { 1 }{ 3 }  }{ 1-\cfrac { 1 }{ 3 }  } $
$=\cfrac { \cfrac { 1 }{ 3 }  }{ \cfrac { 2 }{ 3 }  } =\cfrac { 1 }{ 2 } $
$\therefore { 9 }^{ \cfrac { 1 }{ 3 }  }\times { 9 }^{ \cfrac { 1 }{ 9 }  }\times { 9 }^{ \cfrac { 1 }{ 27 }  }\times ...\infty ={ 9 }^{ \cfrac { 1 }{ 2 }  }$
$=(3^{ 2 })^{ \cfrac { 1 }{ 2 }  }$
$=3$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The sum of the terms of an infinitely decreasing G.P. is $S$. The sum of the squares of the terms of the progression is -

  1. $\dfrac{S}{{2S - 1}}$
  2. $\dfrac{{{S^2}}}{{2S - 1}}$
  3. $\dfrac{S}{{2 - S}}$
  4. ${S^2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$1,r,r^2,r^3..........,(r<0)$


$S _{\infty}=\dfrac {a}{1-r}$

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


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

$S _{\infty}=1^2+r^2+r^4+r^6+..........$

$S _{\infty}=\dfrac {1}{1-r^2}$

       $=\dfrac {1}{1- \left (\dfrac {S-1}{S}\right )^2}$

       $=\dfrac {S^2}{S^2-S^2-1+2S}$

        $=\dfrac {S^2}{2S-1}$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

In a GP the product of the first four terms is 4 and the second term is the reciprocal of the fourth term. The sum of the GP up to infinite terms is-

  1. $2$
  2. $\dfrac{2}{3}$
  3. $-2$

  4. 6
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Let the four terms be $\dfrac { a }{ { r }^{ 2 } } ,\dfrac { a }{ r } ,a,a{ r }$.
Therefore, $\frac { a }{ { r }^{ 2 } } *\frac { a }{ r } *a*a{ r }=4$
$\Rightarrow \dfrac{a^{ 4 }}{r^2}=4$
Also, given that second term is a reciprocal of the fourth term.
So, $\dfrac{a}{ r }=\dfrac { 1 }{ ar } $
$a=\pm 1, r=\pm \dfrac{1}{2}$
Since the sum of infinite terms of G.P is given by $\frac { a }{ 1-r } $. So, by substituting the values of $a$ and $r$; we get
$\frac { a }{ 1-r } =\pm 2$ or $\pm \dfrac{2}{3}$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

Sum to infinity of a G.P is $15$, whose first term is $a$ then a MUST satisfy the inequality given by

  1. $0< a< 130$
  2. $0< a< 30$
  3. $0< a< 15$
  4. $0< a< 100$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

given $a+ar+ar^{2}+ar^{3}+.......=15$ 

where $0 < r < 1  \dfrac{a}{1-r}=15$ 
$(\because |r|\geqslant 1$, geometric series is divergent $)$
$a=15(1-r)$
when $0 < r < 1 $
$\Rightarrow -1 < -r < 0.$
$\Rightarrow  0 < 1-r < 1$
$\therefore  0 < 15(1-r) < 15$
$\Rightarrow 0 < a < 15$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $\displaystyle x=\sum _{a=0}^{\infty }a^{n},y=\sum _{a=0}^{\infty }b^{n},z=\sum _{a=0}^{\infty }c^{n}$ Where $a,b,c $ are in A.P and $\displaystyle \left | a \right |<1,\left | b \right |<1,\left | c \right |<1$ then $x,y,z$ are in

  1. H.P

  2. Arithmetic-Geometric progression

  3. A.P

  4. G.P
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $\displaystyle \left | a \right |< 1,\left | b \right |< 1,\left | c \right |< 1,\ \ a,b,c \in A.P$


and $\displaystyle \sum _{n=0}^{\infty }a^{n}=\frac{1}{1-a},\sum _{n=0}^{\infty }b^{n}=\frac{1}{1-b},\sum _{r=0}^{\infty }c^{n}=\frac{1}{1-c}$

$\displaystyle \therefore x=\frac{1}{1-a},y=\frac{1}{1-b},c=\frac{1}{1-c}$

$\displaystyle \Rightarrow a=\frac{x-1}{x},b=\frac{y-1}{y},c=\frac{z-1}{z}$

$\displaystyle \because 2b=a+c \ as \ a,b,c \in A.P$

$\displaystyle 2\left ( \frac{y-1}{y} \right )=\frac{x-1}{x}+\frac{z-1}{z}\Rightarrow \frac{2}{y}=\frac{1}{x}+\frac{1}{z}$

$\displaystyle \Rightarrow x,y,z \in H.P$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The sum of the infinite series, ${ 1 }^{ 2 }-\frac { { 2 }^{ 2 } }{ 5 } +\frac { { 3 }^{ 2 } }{ { 5 }^{ 2 } } -\frac { { 4 }^{ 2 } }{ { 5 }^{ 3 } } +\frac { { 5 }^{ 2 } }{ { 5 }^{ 4 } } -\frac { { 6 }^{ 2 } }{ { 5 }^{ 5 } } +.........$ is :

  1. $\frac { 1 }{ 2 } $
  2. $\frac { 25 }{ 24 } $
  3. $\frac { 25 }{ 54 } $
  4. $\frac { 125 }{ 252 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is an arithmetico-geometric series of the form sum(n^2 * r^(n-1)). The sum can be found using the method of differences or differentiation of geometric series. The result for this specific series is 1/2.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The first term of an infinitely decreasing G.P. is unity and its sum is S. The sum of the squares of the terms of the progression is

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

Let common ratio is $r<1$
Then G.P is $1,r,{ r }^{ 2 },{ r }^{ 3 },...\infty $
$S=1+r+{ r }^{ 2 }+{ r }^{ 3 }+...\infty $
$\displaystyle \Rightarrow S=\frac { 1 }{ 1-r } $
Then G.P formed by squaring the terms 
$1,{ r }^{ 2 },{ r }^{ 4 },{ r }^{ 6 },...\infty $
$\displaystyle { S }'=\frac { 1 }{ 1-{ r }^{ 2 } } =\frac { 1 }{ \left( 1-r \right) \left( 1+r \right)  } =\frac { { S }^{ 2 } }{ 2S-1. } $

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

Find the sum of the infinite geometric series where the beginning term is $-1$ and the common ratio is $\dfrac{1}{2}$.

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

Given that first term $a=-1$ and common ratio is $r=\dfrac{1}{2}$

We know $\text{sum} = \dfrac{a}{1-r}$
$\Rightarrow \text{sum} = \dfrac{-1}{1-\frac{1}{2}}$
$\Rightarrow \text{sum} = \dfrac{-1}{\frac{1}{2}}$
$\Rightarrow \text{sum} = -2$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $S$ is the sum to infinity of a GP, whose first term is $a$, then the sum of the first $ n$  terms is

  1. $\displaystyle S\left ( 1-\frac{a}{S} \right )^{n}$
  2. $\displaystyle S\left [ 1-\left ( 1-\frac{a}{S} \right )^{n} \right ]$
  3. $\displaystyle a\left [ 1-\left ( 1-\frac{a}{S} \right )^{n} \right ]$

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

Let r be the common ration.
GIven, $S _\infty=S=\frac{a}{1-r}$
$\Rightarrow 1-r=\frac aS$
$\Rightarrow r=1-\frac aS$
Now, sum of n terms is given by
$S _n=\dfrac{a(1-r^n)}{1-r}$


       $=\dfrac{a(1-(1-\frac aS)^n)}{1-(1-\frac aS)}$

       $=\dfrac{a(1-(1-\frac aS)^n)}{\frac aS}$


       $=S[1-(1-\frac aS)^n]$
Option B is correct.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

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

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

Sum of GP $=\dfrac{a(r^n -1)}{r-1}$

$a= $ first term
$r=$ common ratio
For $n \to \infty$
$r^n = 0$ for $r<1$
$r^n \to \infty $ for $r>1$
Thus $\text{sum} =\dfrac {a}{r-1}$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If ${S} _{p}$ denote the sum of the series $1+{r}^{p}+{r}^{2p}+..$ upto infinity and ${X} _{p}$ be the sum of the series $1-{r}^{p}+{r}^{2p}-..$ upto infinity then $\left( r\in \left( -1,1 \right) -\left{ 0 \right}  \right)$

  1. ${S} _{p}+{X} _{p}={2X} _{2p}$
  2. ${S} _{p}+{X} _{p}={2S} _{2p}$
  3. ${S} _{p}+{X} _{p}={S} _{2p}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} { S _{ p } }\, \, is\, \, m-1+{ r^{ p } }+{ r^{ 2p } }... \ { X _{ p } }=1-{ r^{ p } }+{ r^{ 2p } } \ { S _{ p } }=\frac { 9 }{ { 1-R } } =\frac { 1 }{ { 1-{ r^{ p } } } }  \ { X _{ p } }=\frac { 1 }{ { 1+{ r^{ p } } } }  \ { S _{ p } }+{ X _{ p } }=\frac { 1 }{ { 1-{ r^{ p } } } } +\frac { 1 }{ { 1+{ r^{ p } } } }  \ =\frac { { 1+{ r^{ p } }+1 } }{ { \left( { 1-{ r^{ p } } } \right) \left( { 1+{ r^{ p } } } \right)  } } =\frac { 2 }{ { 1-{ r^{ 2p } } } }  \ =2{ S _{ p } } \  \end{array}$