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 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

The sum of  $10$  terms of the series  $\left( x + \dfrac { 1 } { x } \right) ^ { 2 } + \left( x ^ { 2 } + \dfrac { 1 } { x ^ { 2 } } \right) ^ { 2 } + \left( x ^ { 3 } + \dfrac { 1 } { x ^ { 3 } } \right) ^ { 2 } + \ldots .$  is

  1. $\left( \dfrac { x ^ { 20 } - 1 } { x ^ { 2 } - 1 } \right) \left( \dfrac { x ^ { 22 } + 1 } { x ^ { 20 } } \right) + 20$
  2. $\left( \dfrac { x ^ { 18 } - 1 } { x ^ { 2 } - 1 } \right) \left( \dfrac { x ^ { 11 } + 1 } { x ^ { 9 } } \right) + 20$
  3. $\left( \dfrac { x ^ { 18 } - 1 } { x ^ { 2 } - 1 } \right) \left( \dfrac { x ^ { 11 } - 1 } { x ^ { 9 } } \right) + 20$

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

The n-th term is (x^n + 1/x^n)^2 = x^(2n) + 1/x^(2n) + 2. The sum is sum(x^(2n)) + sum(1/x^(2n)) + sum(2). These are geometric series. The sum of x^(2n) is x^2(x^20-1)/(x^2-1).