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 arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

the sum of the first n terms of the series ${ 1 }^{ 2 }+{ 2.2 }^{ 2 }+{ 3 }^{ 2 }+{ 2.4 }^{ 2 }+{ 5 }^{ 2 }+{ 2.6 }^{ 2 }....is\frac { n(n+1)^{ 2 } }{ 2 } $ when n is even.wheen n is odd the sum is

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

The series is 1^2 + 2*2^2 + 3^2 + 2*4^2 + 5^2 + 2*6^2 + ... For odd n, the sum follows a specific pattern. By testing n=1, sum=1. By testing n=3, sum=1^2 + 2*2^2 + 3^2 = 1 + 8 + 9 = 18. Formula A: 3*3(4)/2 = 18. This matches.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

The sum of series $\sec^{-1}\sqrt {2}+\sec^{-1}\dfrac {\sqrt {10}}{3}+\sec^{-1}\dfrac {\sqrt {50}}{7}+...+\sec^{-1}\sqrt {\dfrac {(n^{2}+1)(n^{2}-2n+2)}{(n^{2}-n+1)^{2}}}$ 

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

The terms are sec^-1(sqrt(2)), sec^-1(sqrt(10)/3), etc. This simplifies to tan^-1(1) + tan^-1(1/3) + ... which is a telescoping series of arctangents.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Calculate the sum of the given series $1+11+111+1111+11111+.....$ upto $9$ terms: 

  1. $123456789$
  2. $987654321$
  3. $999999999$

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

$S=1+11+111+1111+...$ 9 terms

$\Rightarrow 9S=9+99+999+9999+...$ 9 terms
$\Rightarrow 9S=(10-1)+(100-1)+(1000-1)+(10000-1)+...$ 9 terms
$\Rightarrow9S=[10+100+1000+10000+...$ 9 terms $]-9$
$\Rightarrow9S=[\dfrac{10.(10^{9}-1)}{10-1}]-9$
$\Rightarrow9S=[1111111110]-9$
$\Rightarrow9S=1111111101$
$\therefore S=123456789$

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Find sum of the first $10$ terms of the series:
$(1)(5)+(2)(6)+(3)(7)+(4)(8)+....$ 

  1. $506$
  2. $605$
  3. $572$
  4. $563$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have to find the value of $1 \times 4+2\times 6+3 \times 7+4 \times 8+............$ , upto $10$ terms

$\Rightarrow1\times 5+2\times 6+3\times 7+4\times 8+........=1(1+4)+2(2+4)+3(3+4)+4(4+4)+.......$
$\Rightarrow { (1 }^{ 2 }+{ 2 }^{ 2 }+....+{ 10 }^{ 2 })+4(1+2+...10)$
$\Rightarrow \frac{10 \times 11 \times 21}{6}+4(\frac{10 \times 11}{2})=605$
Therefore the option is $B$

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

Sum of $n$ terms of the series  $5+7+13+31+85+...,$  is

  1. $4n+\dfrac12(3^n-1)$
  2. $8n+\dfrac12(3^n-1)$
  3. $2n+\dfrac12(3^n-1)$

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

The given series is $5+7+13+31+85+.............$

$\Rightarrow 4n+(1+3+9+27+81+..........)$
$\Rightarrow 4n+(1+3+3^{2}+3^{3}+3^{4}+...........)$
$\Rightarrow 4n+\frac{1}{2}(3^{n}-1)$
Therefore option $A$ is correct

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Find out the largest term of the sequence  $\displaystyle \frac{1}{503},\displaystyle \frac{4}{524}, \displaystyle \frac{9}{581}, \displaystyle \frac{16} {692},....$

  1. $\displaystyle \frac{25}{875}$
  2. $\displaystyle \frac{36}{1148}$
  3. $\displaystyle \frac{49}{1529}$
  4. $\displaystyle \frac{64}{2036}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

General term can be written as
$T _{n}=\displaystyle \frac{n^{2}}{500+3n^{3}}$
then, $\displaystyle \frac{dT _{n}}{dn}=\displaystyle \frac{n(1000-3n^{3})}{(500+3n^{3})^{2}}$
For max or min of $T _{n}$,
$\displaystyle \frac{dT _{n}}{dn}=0$
$\therefore n=\left ( \displaystyle \frac{1000}{3} \right )^{1/3}=6.933\approx7$

Hence, $T _{7}$ is the largest term. So largest term in the given sequence is $\displaystyle \frac{49}{1529}$

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

The sixth term of an A.P is equal to 2. The value of the common difference of the A.P which makes the product $a _{1} a _{4} a _{5}$ least is given by 

  1. $\displaystyle \frac {8}{5}$
  2. $\displaystyle \frac {5}{4}$
  3. $\displaystyle \frac {2}{3}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given that sixth term of an A.P is 2.
$\Rightarrow { a } _{ 1 }+5d=2$
Consider, $p={ a } _{ 1 }{ a } _{ 4 }{ a } _{ 5 }$
$\Rightarrow p={ a } _{ 1 }({ a } _{ 1 }+3d)({ a } _{ 1 }+4d)$
Now substitute ${ a } _{ 1 }=2-5d$ in the above equation, we get;
$ p=(2-5d)(2-2d)(2-d)$
Now, solving the parentheses, we get;
$p=2[4-16d+17{ d }^{ 2 }-5{ d }^{ 3 }]$
Let, $S=-5{ d }^{ 3 }+17{ d }^{ 2 }-16d+4$
Now, taking the derivative of S w.r.t $d$, we get;
$S\prime =-15{ d }^{ 2 }+34{ d }-16$
Notice that, for $S\prime=0$, we get;
$d=\dfrac { 2 }{ 3 } ,\dfrac { 8 }{ 5 } $
Now, taking the derivative of $S\prime$ w.r.t $d$, we get;
$S\prime \prime =-30d+34$
At $d=\dfrac { 2 }{ 3 }$, we get;
$S\prime \prime =-20+34$
So, $S\prime \prime=14$ which is positive.
Therefore, $d=\dfrac { 2 }{ 3 }$ gives minimum value.
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The first and last term of an A.P. are $1$ and $11$. If the sum of its terms is $36$, then the number of terms will be

  1. $5$
  2. $6$
  3. $7$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$a=1\ a+(n-1)d=11\ 1+(n-1)d=11\ (n-1)d=10$

Also, Sum $= \cfrac { [a+(a+(n-1)d)] }{ 2 } n$
$\Rightarrow 36=n\times \cfrac { 12 }{ 2 }$
$\Rightarrow n=6$

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.