Quantitative Aptitude

Number Series

2,662 Questions
All First differences and arithmetic progressions 812 Successive multiplication or division 429 Step-wise calculations 349 Squares and cubes based 338 Separate alternate positions 241 Inter-term operations 144 Mixed-operation rules 138 Unconventional pattern solving 74 Fractional and decimal operations 60 Prime number based 54 Digit operations 23

Number Series Questions

Multiple choice

Identify the formula for the $n^{th}$ term of the sequence $54, 18, 6....$
  1. $54\left  (\dfrac{1}{3}\right)^{n-1}$
  2. $6 \left (\dfrac{1}{3}\right)^{n-1}$
  3. $3\left (\dfrac{1}{3}\right)^{n}$
  4. $54 \left (\dfrac{1}{3}\right)^{n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a geometric progression with first term a = 54 and common ratio r = 18/54 = 1/3. The formula for the n-th term is a * r^(n-1), which is 54 * (1/3)^(n-1).

Multiple choice

The sum of $1$ st $n$ terms of the series $\cfrac { { 1 }^{ 2 } }{ 1 } +\cfrac { { 1 }^{ 2 }+{ 2 }^{ 2 } }{ 1+2 } +\cfrac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 1+2+3 } +........$
  1. $\cfrac { n+2 }{ 3 } $
  2. $\cfrac {n( n+2 )}{ 3 } $
  3. $\cfrac {n( n-2 )}{ 3 } $
  4. $\cfrac {n( n-2 )}{ 6 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The n-th term is (Sum of first n squares) / (Sum of first n integers) = [n(n+1)(2n+1)/6] / [n(n+1)/2] = (2n+1)/3. Summing this from 1 to n: (1/3) * [2 * n(n+1)/2 + n] = (1/3) * [n^2 + n + n] = n(n+2)/3.

Multiple choice

Let $A$ be the sum of the first $20$ terms and $B$ be the sum of the first $40$ terms of the series $1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2.6^{2} + .....$ If $B - 2A = 100\lambda$, then $\lambda$ is equal to
  1. $464$
  2. $496$
  3. $232$
  4. $248$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series is 1^2, 2*2^2, 3^2, 2*4^2, 5^2, 2*6^2... The sum A (20 terms) and B (40 terms) can be calculated by grouping odd and even terms. The difference B - 2A simplifies to a known sum formula, yielding 24800, so lambda = 248.

Multiple choice

The sum to $200$ terms of the series $1+4+6+5+11+6+.....$ is
  1. $29,600$
  2. $29,800$
  3. $30,200$

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

The series is 1, 4, 6, 5, 11, 6... This is two interleaved series: (1, 6, 11, ...) and (4, 5, 6, ...). Each has 100 terms. Sum 1 = (100/2)(2*1 + 99*5) = 50(2+495) = 24850. Sum 2 = (100/2)(2*4 + 99*1) = 50(8+99) = 5350. Total = 24850 + 5350 = 30200.

Multiple choice

The sum of the series $(2)^2+2(4)^{2}+3(6)^{2}+....$ upto $10$ terms is
  1. $12100$
  2. $11300$
  3. $11200$
  4. $12300$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The general term of the series is T_n = n * (2n)^2 = 4n^3. The sum of the first 10 terms is therefore 4 times the sum of the cubes of the first 10 natural numbers. Using the formula for the sum of cubes, we get 4 * (10 * 11 / 2)^2 = 4 * 3025 = 12100.

Multiple choice

The sum of the series ${1}^{2}+{3}^{2}+{5}^{2}+...$ to $n$ terms is
  1. $\cfrac{n(n+1)(n+!)}{2}$
  2. $\cfrac{n(2n-1)(2n+1)}{3}$
  3. $\cfrac{{(n-1)}^{2}(2n+1)}{6}$
  4. $\cfrac{{(2n+1)}^{3}}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The kth odd number is 2k - 1, so the required sum is the sum of (2k - 1)^2 from k = 1 to n. Its closed form is n(2n - 1)(2n + 1)/3, which is option B.