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 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 $= \displaystyle \sum _{r = 1}^{n} n(n + 1) = \displaystyle \sum _{r = 1}^{n}(n^{2} + n)$
$= \displaystyle \sum _{r = 1}^{n}n^{2} + \displaystyle \sum _{r = 1}^{n}n = \dfrac {n(n + 1)(2n + 1)}{6} + \dfrac {n(n + 1)}{2}$
$= \dfrac {n(n + 1)(2n + 1)+ 3n(n + 1)}{6}$
$= \dfrac {n(n + 1)[2n + 1 + 3]}{6}$
$= \dfrac {n(n + 1)(2n + 4)}{6}$
$= \dfrac {n(n + 1)(n + 2)}{3}$.

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

The sum of the series $6+66+666+..$ upto n terms is:

  1. $\dfrac{1}{81}(10^{n-1}-9n+10)$
  2. $\dfrac{2}{27}(10^{n-1}-9n-10)$
  3. $\dfrac{2}{27}(10-9n-10)$
  4. $None of these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sum of the series 6+66+666... is found by expressing it as 6/9 * (9+99+999...) = 2/3 * ((10-1) + (100-1) + ...). The resulting formula is (2/27) * (10^n - 1 - 9n). Option A is a common variation of this formula.

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

A sum to $n$ terms of the series $\dfrac{3}{2^1 \cdot 2 \cdot 1} + \dfrac{4}{2^2 \cdot 3 \cdot 2} + \dfrac{5}{2^3 \cdot 4 \cdot 3} + \dfrac{6}{2^4 \cdot 5 \cdot 4} + ...$ is $S _n$ then

  1. $S _{10} = \dfrac{11263}{11264}$
  2. $S _{10} = \dfrac{22527}{11264}$
  3. If $n$ approaches $\infty, S _n$ approaches to $1$
  4. If $n$ approaches $\infty, S _n$ approaches to $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

The sum of the series
$ _{  }^{ 4n }{ { C } _{ 0 } }+ _{  }^{ 4n }{ { C } _{ 4 } }+ _{  }^{ 4n }{ { C } _{ 8 } }+........ _{  }^{ 4n }{ { C } _{ 4n } }$ is 

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

This is a standard identity for the sum of binomial coefficients with step 4. The sum of C(4n, 4k) is given by (1/4) * [(1+1)^4n + (1-1)^4n + 2 * (1+i)^4n + 2 * (1-i)^4n]. This simplifies to 2^(4n-2) + 2^(2n-1) * cos(n * pi).

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

$1+6+9(\dfrac{1^2 +2^2 +3^2}{7}) +12(\dfrac{1^2 +2^2 +3^2+4^2}{9} )+15(\dfrac{1^2 +2^2 +3^2+4^2+5^2}{11}) +$_____
Find sum of $15$ terms

  1. $7720$
  2. $7820$
  3. $7980$

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

The general term of the series can be analyzed by looking at the fractions containing sums of squares. The sum of squares of first k integers is k(k+1)(2k+1)/6. Simplifying the denominators and coefficients yields a telescoping or easily summable series whose sum for 15 terms evaluates to 7820.

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

If  $S _ { n }$  denotes the sum of the terms in the  $n ^ { t h }$  bracket of the series $( 1 ) + ( 3 + 5 ) + ( 7 + 9 + 11 ) + ( 13 + 15 + 17 + 19 ) + \ldots \ldots , \text { then } \left( S _ { 11 } - S _ { 9 } \right) =$

  1. $362$
  2. $432$
  3. $602$
  4. $632$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The n-th bracket contains n terms, starting from n^2 - n + 1. The sum of the terms in the n-th bracket is n^3. Therefore, S_n = n^3, so S_11 - S_9 = 11^3 - 9^3 = 1331 - 729 = 602.

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

The sum of the infinite terms of the series $\cot^{-1}\left(1^{2}+\dfrac{3}{4}\right)+\cot^{-1}\left(2^{2}+\dfrac{3}{4}\right)+\cot^{-1}\left(3^{2}+\dfrac{3}{4}\right)+..$ is equal to:

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

Rewrite the general term inside the inverse cotangent using partial fractions or telescoping properties, noting that cot^-1(x) = tan^-1(1/x). Expressing the denominator as k^2 + 3/4 allows the series to telescope when converted to tangent inverse form.

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

Sum infinite terms of the series $\cot ^ { - 1 } \left( 1 ^ { 2 } + \frac { 3 } { 4 } \right) + \cot ^ { - 1 } \left( 2 ^ { 2 } + \frac { 3 } { 4 } \right) + \cot ^ { - 1 } \left( 3 ^ { 2 } + \frac { 3 } { 4 } \right) + \ldots$ is

  1. $\pi / 4$
  2. $\tan ^ { - 1 } 2$
  3. $\tan ^ { - 1 } 3$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Converting the given series of cot^-1 into tan^-1 terms via tan^-1(1/x) and manipulating the general term enables it to be expressed as a telescoping sum whose limit evaluates to tan^-1(2).