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

Select the missing number from the given series. $36, 18, 54, 27, 81, ?, 121.5$

  1. 40

  2. 38

  3. 40.5

  4. 41.5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Find the pattern in the series: 36, 18, 54, 27, 81, ?, 121.5. Pattern: alternate operations - divide by 2, then multiply by 3. 36÷2=18, 18×3=54, 54÷2=27, 27×3=81, 81÷2=40.5, 40.5×3=121.5. The missing number is 81÷2=40.5. Options A(40), B(38), and D(41.5) are incorrect.

Multiple choice

The sum to $2n$ terms of the series $\displaystyle \log _{3}1-\log _{3}3+\log _{3}9-\log _{3}27....$ is
  1. $\displaystyle \frac{\left ( 2n-1 \right )}{\left ( n-1 \right )}$
  2. $\displaystyle \frac{\left ( n-1 \right )}{\left ( 2n-1 \right )}$
  3. $1 - n$
  4. $-n$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series is log3(1) - log3(3) + log3(9) - log3(27) + ... which simplifies to 0 - 1 + 2 - 3 + ... . This is an alternating sum of integers. The sum of the first 2n terms is (0-1) + (2-3) + ... + ((2n-2) - (2n-1)), which is -1 + -1 + ... + -1 (n times). Thus, the sum is -n.

Multiple choice

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

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

The general term is n / (1 + n^2 + n^4). Note that 1 + n^2 + n^4 = (n^2 + n + 1)(n^2 - n + 1). Using partial fractions, the sum telescopes to (n^2 + n) / (2(n^2 + n + 1)).

Multiple choice

The sum of $n$ terms of the series $1+\dfrac { 4 }{ 5 } +\dfrac { 7 }{ { 5 }^{ 2 } } +\dfrac { 10 }{ { 5 }^{ 3 } } +\cdots $ is
  1. $\dfrac { 5 }{ 4 } +\dfrac { 15 }{ 16 } \left( 1-\dfrac { 1 }{ { 5 }^{ n-1 } } \right) -\dfrac { \left( 3n-2 \right) }{ 4\cdot { 5 }^{ n-1 } } $
  2. $\dfrac { 5 }{ 4 } +\dfrac { 1 }{ 16 } \left( 1-\dfrac { 1 }{ { 5 }^{ n-1 } } \right) -\dfrac { 3n }{ 4\cdot { 5 }^{ n-1 } } $
  3. $\left( 1-\dfrac { 1 }{ { 5 }^{ n-1 } } \right) -\dfrac { \left( 3n+2 \right) }{ 4\cdot { 5 }^{ n-1 } } $

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

This is an arithmetico-geometric series. The sum of the first n terms follows the standard formula for such series.

Multiple choice

Find the sum to $n$ terms of the sequence $$ given by $a_{n} = 2^{n} + 3n, n\epsilon N$.
  1. $2(2^{n} + 1) + \dfrac {3n}{2}(n + 1)$
  2. $2(2^{n} - 1) + \dfrac {3n}{2}(n + 1)$
  3. $(2^{n} - 1) + \dfrac {3n}{2}(n + 1)$

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

The sum is the sum of a geometric series (2^n) and an arithmetic series (3n). Sum(2^k) from 1 to n is 2(2^n - 1). Sum(3k) from 1 to n is 3n(n+1)/2. Adding these gives 2(2^n - 1) + 3n(n+1)/2.

Multiple choice

The sum of the first $20$ terms of the series $1+\displaystyle\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+....$, is?
  1. $\displaystyle 38+\frac{1}{2^{20}}$
  2. $\displaystyle 39+\frac{1}{2^{19}}$
  3. $\displaystyle 39+\frac{1}{2^{20}}$
  4. $\displaystyle 38+\frac{1}{2^{19}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The series is 1 + (1 + 1/2) + (1 + 3/4) + (1 + 7/8) + ... = sum_{n=0}^{19} (1 + (2^n - 1)/2^n) = sum (2 - 1/2^n) = 2*20 - sum_{n=0}^{19} (1/2^n). Sum of GP = (1 - (1/2)^20) / (1 - 1/2) = 2(1 - 1/2^20) = 2 - 1/2^19. Total = 40 - (2 - 1/2^19) = 38 + 1/2^19.

Multiple choice

If the sum of the first ten terms of the series $\left (1 \dfrac {3}{5}\right )^{2} + \left (2 \dfrac {2}{5}\right )^{2} + \left (3 \dfrac {1}{5}\right )^{2} + 4^{2} + \left (4 \dfrac {4}{5}\right )^{2} + ..............$ is $\dfrac {16}{5}m$, then $m$ is equal to:
  1. $102$
  2. $101$
  3. $100$
  4. $99$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Series terms: (8/5)^2, (12/5)^2, (16/5)^2, (20/5)^2... This is (4/5)^2 * (2^2, 3^2, 4^2, 5^2...). The sum of the first 10 terms is (16/25) * sum of squares from 2 to 11. Sum of squares 1^2 to 11^2 = (11*12*23)/6 = 506. Subtract 1^2 = 505. Sum = (16/25) * 505 = 16 * 20.2 = 323.2. 323.2 = (16/5) * m. m = 323.2 * 5 / 16 = 101.

Multiple choice

The sum of first $9$ terms of the series $\displaystyle \dfrac {1^3}{1}+\dfrac {1^3+2^3}{1+3}+\dfrac {1^3+2^3+3^3}{1+3+5}+ ....$ is
  1. $71$
  2. $96$
  3. $142$
  4. $192$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The n-th term is (1^3 + ... + n^3) / (1 + 3 + ... + (2n-1)). Sum of cubes = (n(n+1)/2)^2. Sum of odd numbers = n^2. Term = (n^2(n+1)^2/4) / n^2 = (n+1)^2 / 4. Sum = 1/4 * sum of (n+1)^2 for n=1 to 9. Sum = 1/4 * (2^2 + 3^2 + ... + 10^2) = 1/4 * (385 - 1) = 96.