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 $n$th term of the series $5+7+13+31+85+\cdots$ is
  1. ${ 3 }^{ n-1 }+2$
  2. $3+{ 3 }^{ n-1 }$
  3. $4+{ 3 }^{ n-1 }$

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

The series is 5, 7, 13, 31, 85. The differences are 2, 6, 18, 54. These are 2 * 3^0, 2 * 3^1, 2 * 3^2, 2 * 3^3. The nth term is 5 + sum(2 * 3^(k-1)) for k=1 to n-1. This equals 5 + 2 * (3^(n-1) - 1) / (3 - 1) = 5 + 3^(n-1) - 1 = 4 + 3^(n-1).

Multiple choice

The sum of first two terms of an infinite G.P. is $5$ and each term is three times the sum of following terms, then the 4th term of the series is $____.$
  1. $1$
  2. $\dfrac 14$
  3. $\dfrac 1{16}$
  4. $\dfrac 18$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the G.P. be a, ar, ar^2... Sum of first two = a(1+r) = 5. Each term is 3 times the sum of following terms: a*r^n = 3 * (a*r^(n+1) / (1-r)). This simplifies to 1 = 3r / (1-r), so 1-r = 3r, 4r = 1, r = 1/4. Then a(1 + 1/4) = 5, a(5/4) = 5, a = 4. The 4th term is ar^3 = 4 * (1/4)^3 = 4 / 64 = 1/16.

Multiple choice

Find the sum to $n$ terms of the series $0.7+7.7+0.77+77.7+0.777+777.7+0.7777+....$ where $n$ is even.
  1. $\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { n }{ 4 } -\displaystyle \frac { \left( 1-\left( 0.1 \right) ^{ n/4 } \right) }{ 9 } \right\} +\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { 10((10)^{ n/4 }-1) }{ 9 } -0.1\displaystyle \frac { n }{ 4 } \right\} $
  2. $\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { n }{ 2 } -\displaystyle \frac { \left( 1-\left( 0.1\right) ^{ n/2 } \right) }{ 9 } \right\} +\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { 10((10)^{ n/2 }-1) }{ 9 } -\displaystyle \frac { n }{ 2 } \right\} $
  3. $\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { n }{ 2 } +\displaystyle \frac { \left( 1-\left( 0.1 \right) ^{ n/2 } \right) }{ 9 } \right\} +\displaystyle \frac { 7 }{ 9 } \left\{\displaystyle \frac { 10((10)^{ n/2 }-1) }{ 9 } +\displaystyle \frac { n }{ 2 } \right\} $

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

The series is (0.7 + 0.77 + ... n/2 terms) + (7.7 + 77.7 + ... n/2 terms). This is a complex summation problem involving geometric series components. Option B correctly represents the sum of these two distinct parts.