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

Find ${a}_{30}$ given that the first few terms of a geometric sequence are given by $-2,1,-\dfrac {1}{2},\dfrac {1}{4}....$
  1. $\dfrac {1}{2^{27}}$
  2. $\dfrac {1}{2^{26}}$
  3. $\dfrac {1}{2^{28}}$
  4. $\dfrac {1}{2^{29}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sequence is -2, 1, -1/2, 1/4. This is a geometric progression with first term a = -2 and common ratio r = -1/2. The n-th term is a * r^(n-1). For n=30, term = -2 * (-1/2)^29 = -2 * (-1 / 2^29) = 2 / 2^29 = 1 / 2^28.

Multiple choice

You are given a sequence of $58$ terms; each term has the form $P + n$ where $P$ stands for the product $2.3.5...61$ of all prime numbers (a prime number is a number divisible only by $1$ and itself) less than or equal to $61$, and $n$ takes, successively, the value $2, 3, 4, ..., 59$. Let $N$ be the number of primes appearing in this sequence. Then $N$ is
  1. $0$
  2. $16$
  3. $17$
  4. $57$
  5. $58$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sequence terms are P+2, P+3, ..., P+59. Since P is the product of all primes up to 61, each term P+n is divisible by n for 2 <= n <= 59. Thus, none of these numbers can be prime.

Multiple choice

The sum of $n$ terms of the following series; $1^3 + 3^3 + 5^3 + 7^3 + ....$ is
  1. $n^2(2n^2-1)$
  2. $n^3(n-1)$
  3. $n^3 + 8n + 4$
  4. $2n^4 + 3n^2$
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 k cubes is (k(k+1)/2)^2. The sum of the first 2n integers is (2n(2n+1)/2)^2 = n^2(2n+1)^2. The sum of even cubes (2^3 + 4^3 + ... + (2n)^3) is 8 * (1^3 + 2^3 + ... + n^3) = 8 * (n(n+1)/2)^2 = 2n^2(n+1)^2. Subtracting even from total: n^2(2n+1)^2 - 2n^2(n+1)^2 = n^2(4n^2 + 4n + 1 - 2n^2 - 4n - 2) = n^2(2n^2 - 1).

Multiple choice

The $n$th term of the series $1+2+5+12+25+\cdots $ is
  1. $\left( n-1 \right) \left( n-2 \right) $
  2. $\dfrac { 1 }{ 3 } n\left( n-1 \right) \left( n-2 \right) +n$
  3. $n$

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

The series is 1, 2, 5, 12, 25. Differences are 1, 3, 7, 13. Second differences are 2, 4, 6. This indicates a cubic sequence. Testing n=1, 2, 3 gives the correct values for the formula 1/3 * n(n-1)(n-2) + n.

Multiple choice

The sum of $n$ terms of the series $\sum _{ x=1 }^{ n }{ \log { \cfrac { { 2 }^{ x } }{ { 3 }^{ x-1 } } } } $ is
  1. $\log { { \left( \cfrac { { 2 }^{ n-1 } }{ { 3 }^{ n+1 } } \right) }^{ n/2 } } $
  2. $\log { { \left( \cfrac { { 2 }^{ n+1 } }{ { 3 }^{ n-1 } } \right) }^{ n/2 } } $
  3. $\log { { \left( \cfrac { { 3 }^{ n-1 } }{ { 2 }^{ n+1 } } \right) }^{ n/2 } } $
  4. $\log { { \left( \cfrac { { 3 }^{ n+1 } }{ { 2 }^{ n-1 } } \right) }^{ n/2 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Sum = log(2^1/3^0) + log(2^2/3^1) + ... + log(2^n/3^(n-1)) = log( (2^1 * 2^2 * ... * 2^n) / (3^0 * 3^1 * ... * 3^(n-1)) ). Numerator exponent = n(n+1)/2. Denominator exponent = (n-1)n/2. Sum = log( 2^(n(n+1)/2) / 3^(n(n-1)/2) ). This simplifies to log( (2^(n+1) / 3^(n-1))^(n/2) ).

Multiple choice

If ${n^{th}}$ term of the series $25 + 29 + 33 + 37 + ....$ and $3+4+6+9+13+....$ are equal, the $n$ equals
  1. $11$
  2. $12$
  3. $13$
  4. $14$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The n-th term of the first series is 25 + (n-1)4 = 4n + 21. The second series is 3, 4, 6, 9, 13, which has second differences of 1, so the n-th term is a quadratic of the form an^2 + bn + c, specifically 0.5n^2 - 0.5n + 3. Setting 4n + 21 = 0.5n^2 - 0.5n + 3 leads to n^2 - 9n - 36 = 0, which factors to (n-12)(n+3) = 0, giving n = 12.

Multiple choice

Find the sum of the series $3+7+14+24+37+ ...... $up to $10$ terms.
  1. $570$
  2. $750$
  3. $705$
  4. $507$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The differences between consecutive terms of the series are 4, 7, 10, 13, which form an arithmetic progression with a first term of 4 and a common difference of 3. The general term of the series is a_n = 1.5*n^2 - 0.5*n + 2. Summing this expression from n = 1 to 10 yields 1.5*385 - 0.5*55 + 20 = 570.

Multiple choice

If the first three terms of an arithmetic sequence are $a,2a,18a-5b$ and further it is known that $a+b+4=0,$ then the value of the $100th$ term of the sequence would be?
  1. $-100$
  2. $-300$
  3. $150$
  4. $-150$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an arithmetic sequence, the difference between consecutive terms is constant: 2a - a = (18a - 5b) - 2a. This simplifies to a = 16a - 5b, or 15a = 5b, so b = 3a. Substituting into a + b + 4 = 0 gives a + 3a + 4 = 0, so 4a = -4 and a = -1. The sequence is -1, -2, -3, where the n-th term is -n, making the 100th term -100.