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 the first $25$ terms of an arithmetic sequence is $1,400$, and the $25th$ term is $104$. F ind the value of $a_{2} - a_{1}$ where the first term of the sequence is $a_{1}$ and the second term is $a_{2}$.
  1. $-3$
  2. $2$
  3. $4$
  4. $5$
  5. $8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sum of an arithmetic sequence is given by S_n = (n/2)(a_1 + a_n). Substituting the given values, 1400 = (25/2)(a_1 + 104), which simplifies to a_1 = 8. Using the formula for the 25th term, 104 = 8 + 24*d, we find the common difference d = a_2 - a_1 = 4.

Multiple choice

The first term of a sequence of numbers is $-3$. Each term after the first is obtained by multiplying the preceding term by $-1$ and then subtracting $1$. What is the $75^{th}$ term of the sequence?
  1. $-73$
  2. $-3$
  3. $2$
  4. $4$
  5. $73$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The sequence is defined by a_1 = -3 and a_(n+1) = -a_n - 1. Calculating the first few terms: a_1 = -3, a_2 = -(-3) - 1 = 2, a_3 = -(2) - 1 = -3, a_4 = -(-3) - 1 = 2. The sequence alternates between -3 and 2, where odd-indexed terms are -3 and even-indexed terms are 2.

Multiple choice

The sum of the first n terms of the series ${ 1 }^{ 2 }+{ 2.2 }^{ 2 }+{ 3 }^{ 2 }+{ 2.4 }^{ 2 }+{ 5 }^{ 2 }+{ 2.6 }^{ 2 }+.....$ is $\cfrac { n{ \left( n+1 \right) }^{ 2 } }{ 2 } $, when n is even. When n is odd, the sum is-
  1. $\cfrac { { n }^{ 2 }(n+1) }{ 2 } $
  2. $\cfrac { n(n+1)(2n+1) }{ 6 } $
  3. $\cfrac { n{ (n+1) }^{ 2 } }{ 2 } $
  4. $\cfrac { { n }^{ 2 }{ (n+1) }^{ 2 } }{ 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series is 1^2 + 2.2^2 + 3^2 + 2.4^2 + 5^2 + 2.6^2... For odd n, the sum follows the pattern n^2(n+1)/2.

Multiple choice

The sum to $n$ terms of the series $\displaystyle \frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+...$ is
  1. $\displaystyle \frac{6n}{n+1}$
  2. $\displaystyle \frac{9n}{n+1}$
  3. $\displaystyle \frac{12n}{n+1}$
  4. $\displaystyle \frac{3n}{n+1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The nth term is (2n+1) / (sum of first n squares). The sum of first n squares is n(n+1)(2n+1)/6. Thus, the nth term is (2n+1) / (n(n+1)(2n+1)/6) = 6 / (n(n+1)) = 6(1/n - 1/(n+1)). Summing this telescoping series gives 6(1 - 1/(n+1)) = 6n/(n+1).