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

Sum to $n$ terms of the series $\dfrac { 1 }{ 1.2.3.4 } +\dfrac { 1 }{ 2.3.4.5 } +\dfrac { 1 }{ 3.4.5.6 } +..........$, is
  1. $\dfrac { { n }^{ 3 } }{ 3\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }$
  2. $\dfrac { { n }^{ 3 }+{ 6n }^{ 2 }-3n }{ 6\left( n+2 \right) \left( n+3 \right) \left( n+4 \right) }$
  3. $\dfrac { 15{ n }^{ 2 }+7n }{ 4n\left( n+1 \right) \left( n+5 \right) }$
  4. $\dfrac { { n }^{ 3 }+6{ n }^{ 2 }+11n }{ 18\left( n+1 \right) \left( n+2 \right) \left( n+3 \right) }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The general term of the series is 1/(n(n+1)(n+2)(n+3)). Using the method of differences, this can be written as (1/3) * [1/(n(n+1)(n+2)) - 1/((n+1)(n+2)(n+3))]. Summing this telescoping series leads to the expression in option D.

Multiple choice

The sum of the series $\dfrac {3}{1^{2}} + \dfrac {5}{1^{2} + 2^{2}} + \dfrac {7}{1^{2} +2^{2} + 3^{2}} + ....$ upto $n$ terms, is
  1. $\dfrac {n}{n + 1}$
  2. $\dfrac {n + 2}{n + 1}$
  3. $\dfrac {6n}{n + 1}$
  4. $\dfrac {6(n + 2)}{n + 1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The n-th term is (2n+1) / (sum of first n squares). Sum of first n squares = n(n+1)(2n+1)/6. So n-th term = (2n+1) / [n(n+1)(2n+1)/6] = 6 / [n(n+1)] = 6 * (1/n - 1/(n+1)). The sum of this telescoping series is 6 * (1 - 1/(n+1)) = 6 * (n / (n+1)) = 6n / (n+1).

Multiple choice

The sum of the series $1^{3}+3^{3}+5^{3}+....$ upto $20$ terms is
  1. $319600$
  2. $321760$
  3. $306000$
  4. $347500$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series is 1^3, 3^3, 5^3, ... The n-th term is (2n-1)^3. The sum is sum_{n=1}^{20} (8n^3 - 12n^2 + 6n - 1). Using summation formulas: 8 * (20*21/2)^2 - 12 * (20*21*41/6) + 6 * (20*21/2) - 20 = 8 * 44100 - 12 * 2870 + 1260 - 20 = 352800 - 34440 + 1240 = 319600.

Multiple choice

The number of common terms between the sequences given by $1,4,7,10,.....298$ and $2,4,6,8,.....300$ is
  1. $49$
  2. $50$
  3. $48$
  4. $100$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sequence 1: 1, 4, 7, ..., 298 (Arithmetic, a=1, d=3). Sequence 2: 2, 4, 6, ..., 300 (Arithmetic, a=2, d=2). Common terms must be even and in sequence 1. Sequence 1 terms: 3n-2. 3n-2 must be even, so 3n must be even, n must be even. n=2, 4, ..., 100. Terms are 4, 10, 16, ..., 298. This is an AP with a=4, d=6. 298 = 4 + (k-1)6 => 294 = (k-1)6 => 49 = k-1 => k=50.