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

The $n^{th}$ terms of the series $3 + 7 + 13 + 21 + \dots $ is
  1. $4n - 1$
  2. $n^n + 2n$
  3. $n^2 + n + 1$
  4. $n^2 + 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sequence is 3, 7, 13, 21. The differences are 4, 6, 8, which form an arithmetic progression. The n-th term is a quadratic of the form an^2 + bn + c. Solving for n=1, 2, 3 gives 3, 7, 13, which yields n^2 + n + 1.

Multiple choice

The sum of $40$ terms of the series $1+2+3+4+5+8+7+16+9+......$ is,
  1. $398 + 2^{20}$
  2. $398 + 2^{21}$
  3. $398 + 2^{19}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The series is 1, 2, 3, 4, 5, 8, 7, 16, 9, 32... The odd terms are 1, 3, 5, 7, 9... (arithmetic progression). The even terms are 2, 4, 8, 16, 32... (geometric progression). Sum of 20 odd terms = 20^2 = 400. Sum of 20 even terms = 2(2^20 - 1) / (2 - 1) = 2^21 - 2. Total sum = 400 + 2^21 - 2 = 398 + 2^21.

Multiple choice

Sum to $n$ terms of the series $ \displaystyle \frac {1}{1.2.3}+\frac {3}{2.3.4}+\frac {5}{3.4.5}+\frac {7}{4.5.6}+ ....$is
  1. $\displaystyle \frac {n(n+1)}{2(n+2)(n+3)}$
  2. $\displaystyle \frac {n(3n+5)}{4(n+1)(n+2)}$
  3. $\displaystyle \frac {1}{6}-\frac {5}{(n+1)(n+4)}$
  4. $\displaystyle \frac {n(3n+1)}{4(n+1)(n+2)}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The general term is a_n = (2n-1) / (n(n+1)(n+2)). Using partial fractions, this simplifies to 1/2 * [1/(n(n+1)) - 1/((n+1)(n+2))]. Summing this telescoping series leads to the result n(3n+1) / (4(n+1)(n+2)).

Multiple choice

The sum to $(n+1)$ terms of the series $\displaystyle \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} +\cdots$ is
  1. $ \dfrac{1}{n+1} $
  2. $ \dfrac{1}{n+2} $
  3. $ \dfrac{1}{n(n+1)} $
  4. $ \dfrac{1}{(n+1)(n+2)} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series is a sum of terms of the form ((-1)^k * C_k) / (k+2). Using the identity integral from 0 to 1 of x^k * (1-x)^n dx = n! * k! / (n+k+1)!, one can evaluate this sum. The result is 1 / ((n+1)(n+2)).

Multiple choice

The sum of series $\displaystyle \frac {1^2}{1}+\frac {1^2+2^2}{1+2}+\frac {1^2+2^2+3^2}{1+2+3}+....$ upto n terms is
  1. $\displaystyle\frac {1}{3}(2n+1)$
  2. $\displaystyle\frac {1}{3}n^2$
  3. $\displaystyle\frac {1}{3}(n+2)$
  4. $\displaystyle\frac {1}{3}n(n+2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice

Sum the series $1^3+3^3+5^3+.......... $ to $n$ terms is
  1. $ n^2\left ( 2n^2-1 \right )$
  2. $n\left ( 2n^2-1 \right )$
  3. $n\left ( 2n^2+1 \right )$
  4. $n^2\left ( 2n^2+1 \right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series is 1^3 + 3^3 + 5^3 + ... + (2n-1)^3. The sum of the first (2n-1) cubes is [ (2n-1)2n/2 ]^2 = n^2(2n-1)^2. Subtracting the even cubes (2^3 + 4^3 + ... + (2n-2)^3) = 8 * [1^3 + 2^3 + ... + (n-1)^3] = 8 * [ (n-1)n/2 ]^2 = 2n^2(n-1)^2. Result is n^2(4n^2 - 4n + 1 - 2n^2 + 4n - 2) = n^2(2n^2 - 1).

Multiple choice

The sum upto $(2n+1)$ terms of the series $a^2-(a+d)^2+(a+2d)^2-(a+3d)^2+ ....$ is
  1. $a^2+3nd^2$
  2. $a^2+2nad+n(n-1)d^2$
  3. $a^2+3nad+n(n-1)d^2$
  4. $a^2+2nad+n(2n+1)d^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series is a^2 - (a+d)^2 + (a+2d)^2 - (a+3d)^2 + ... + (a+2nd)^2. Grouping terms: (a^2 - (a+d)^2) + ((a+2d)^2 - (a+3d)^2) + ... + ((a+(2n-2)d)^2 - (a+(2n-1)d)^2) + (a+2nd)^2. Each pair is of the form x^2 - y^2 = (x-y)(x+y) = -d(2a + (4k+1)d). Summing these leads to the expression a^2 + 2nad + n(2n+1)d^2.