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 sum of series $\displaystyle {1 \over 2} + {3 \over 4} + {7 \over 8} + {{15} \over {16}} + ...$ up to n term is-
  1. $n - 1 + {2^{ - n}}$
  2. ${2^{ - n}} - 1$
  3. $n - 1$

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

The series is (1 - 1/2) + (1 - 1/4) + (1 - 1/8) + ... + (1 - 1/2^n). Sum = n - (1/2 + 1/4 + ... + 1/2^n). The sum of the geometric series 1/2 + 1/4 + ... + 1/2^n is (1/2 * (1 - (1/2)^n)) / (1 - 1/2) = 1 - (1/2)^n. Total sum = n - (1 - 1/2^n) = n - 1 + 2^-n.

Multiple choice

The number of terms of an $A.P.$ is even; the sum of the odd terms is $24$, and of the even terms is $30$ and the last term exceeds the first by $10.5$, then the number of terms in the series is
  1. $8$
  2. $4$
  3. $6$`
  4. $10$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the series contain 2m terms. The difference between the sums of even and odd-positioned terms is md, so md = 6. The last-minus-first condition gives (2m - 1)d = 10.5, yielding m = 4 and therefore 8 terms.

Multiple choice

The $nth$ term of the series $4, 14, 30, 52, 80, 114, ...$ is
  1. $n^{2} + n + 2$
  2. $3n^{2} + n$
  3. $3n^{2} - 5n + 2$
  4. $(n + 1)^{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Series: 4, 14, 30, 52, 80, 114. Differences: 10, 16, 22, 28, 34. Second differences: 6, 6, 6, 6. This is a quadratic sequence an^2 + bn + c. 2a = 6 => a = 3. 3a + b = 10 => 9 + b = 10 => b = 1. a + b + c = 4 => 3 + 1 + c = 4 => c = 0. Formula: 3n^2 + n.

Multiple choice

The sum to n terms of the series $3^2 + 8^2 + 13^2$ +.......
  1. $\dfrac{n(50n^2 + 15n-11)}{6}$
  2. $\dfrac{(50n^2+15n-11)}{6}$
  3. $\dfrac{n(50n^2+15n+11)}{6}$
  4. $\dfrac{n(50n^2-15n-11)}{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series is sum of (5k-3)^2 for k=1 to n. (5k-3)^2 = 25k^2 - 30k + 9. Sum = 25 * [n(n+1)(2n+1)/6] - 30 * [n(n+1)/2] + 9n. Simplifying this leads to n(50n^2 + 15n - 11) / 6.

Multiple choice

Consider two arithmetic series : $\begin{array} { l } { A _ { 1 } : 2 + 9 + 16 + 23 + \ldots \ldots \ldots + 205 } \ { A _ { 2 } : 5 + 9 + 13 + 17 + \ldots \ldots \ldots + 161 } \end{array}$ then the number of terms common to the two series is
  1. $6$
  2. $8$
  3. $10$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A1: 2, 9, 16, ..., 205 (a=2, d=7). A2: 5, 9, 13, ..., 161 (a=5, d=4). Common terms start at 9, with common difference = LCM(7,4) = 28. Series: 9, 37, 65, 93, 121, 149. Next term 177 > 161. Total 6 terms.