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 numbers and sequences series introduction to series introduction to sequences and series

Sum the following series to n terms: $3+5+9+15+23+...$

  1. $\dfrac{n}{3}(n^{2}-8)$
  2. $\dfrac{n}{3}(n^{2}+8)$
  3. $\dfrac{n}{2}(n^{3}+8)$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We have 

$3+5+9+15+23+............+T _{n-1}+T _{n}$

The difference between the successive terms are $S$ difference are in $A.P$

Let $S _{n}$ denote the sum to $n$ terms of the given series 

Then, $S _{n}=3+5+9+15+23+........+ T _{n-1}+T _{n}..... (1)$

Also $S _{n}=3+5+9+15+.......+T _{n-1}+T _{n}........ (2)$

Subtracting $(2)$ from $(1)$ we get 

$0=3+\dfrac{(n-1)}{2}[2\times 2+(n-1-1)\times 2)]$

$T _{n}=3+(n-1)(n)$

$=3+n^{2}-n=n^{2}-n+3$

$S _{n}=\displaystyle\sum _{k=1}^{n}k^{2}=\sum _{k=1}^{n}k+3\sum _{k=1}^{n}1$

$=\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n(n+1)}{2}+3n$

$=\dfrac{n(n+1)(2n+1)-3n(n+1)+18n}{6}$

$=\dfrac{n}{6}[2n^{2}+3n-3n-2+18]=\dfrac{n}{6}[2n^{2}+16]$

$=\dfrac{n}{6}\times 2[n^{2}+8]=\dfrac{n}{3}[n^{2}+8]$

Hence $S _{n}=\dfrac{n}{3}(n^{2}+8)$
Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

Find the first five terms of the sequence specified by the recursion formula
${a} _{k+1}={a} _{k}+3$, if ${a} _{1}=7$.

  1. $7,10,13,16,19$
  2. $6,9,12,15,18$
  3. $8,11,14,17,20$
  4. $5,8,11,14,17$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ${a} _{k+1} = {a} _{k}+3$ and ${a} _{1}=7$

So ${a} _{2}={a} _{1}+3 = 7+3=10$
${a} _{3}={a} _{2}+3 = 10+3=13$
${a} _{4}={a} _{3}+3 = 13+3=16$
${a} _{5}={a} _{4}+3=16+3=19$

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

The sum of the series $1-\cfrac { 3 }{ 2 } +\cfrac { 5 }{ 4 } -\cfrac { 7 }{ 8 } +...\infty $ is

  1. $\cfrac { 2 }{ 9 } $
  2. $\cfrac { -4 }{ 9 } $
  3. $\cfrac { 4 }{ 9 } $
  4. $\cfrac { -2 }{ 9 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$S=1-\cfrac { 3 }{ 2 } +\cfrac { 5 }{ 4 } -\cfrac { 7 }{ 8 } +...\infty $
$\cfrac { S }{ 2 } =\cfrac { 1 }{ 2 } -\cfrac { 3 }{ 4 } +\cfrac { 5 }{ 8 } $
$S\left( 1+\cfrac { 1 }{ 2 }  \right) =1-\cfrac { 2 }{ 2 } +\cfrac { 2 }{ 4 } -\cfrac { 2 }{ 8 } ....\infty $
$S\left( \cfrac { 3 }{ 2 }  \right) =1-1+\cfrac { 1 }{ 2 } -\cfrac { 1 }{ 4 } ....\infty $
$S\left( \cfrac { 3 }{ 2 }  \right) =\cfrac { 1 }{ 2 } \left( 1-\cfrac { 1 }{ 2 } ....\infty  \right) $
$S\left( \cfrac { 3 }{ 2 }  \right) =\cfrac { 1 }{ 2 } \left( \cfrac { 1 }{ 1+1/2 }  \right) $
$S\left( \cfrac { 3 }{ 2 }  \right) =\cfrac { 1 }{ 2 } .\cfrac { 2 }{ 3 } $
$S\left( \cfrac { 3 }{ 2 }  \right) =\cfrac { 1 }{ 3 } $
$S=\cfrac { 2 }{ 9 } $
Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

$1\times 2 + 2\times 3 + 3\times 4 + .... n\ terms =$

  1. $\dfrac {n(n + 1)(n + 2)}{3}$
  2. $\dfrac {n(n + 1)(n - 2)}{3}$
  3. $\dfrac {n(n + 1)(n + 2)}{6}$
  4. $\dfrac {n(n - 1)(n - 2)}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$n^{th}$ term of sequence $= n(n + 1)$
Then sum of $n$ terms $= \displaystyle \sum _{r = 1}^{n} n(n + 1) = \displaystyle \sum _{r = 1}^{n}(n^{2} + n)$
$= \displaystyle \sum _{r = 1}^{n}n^{2} + \displaystyle \sum _{r = 1}^{n}n = \dfrac {n(n + 1)(2n + 1)}{6} + \dfrac {n(n + 1)}{2}$
$= \dfrac {n(n + 1)(2n + 1)+ 3n(n + 1)}{6}$
$= \dfrac {n(n + 1)[2n + 1 + 3]}{6}$
$= \dfrac {n(n + 1)(2n + 4)}{6}$
$= \dfrac {n(n + 1)(n + 2)}{3}$.