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

What is the 17th term of the sequence 3, 6, 11, 18, 27, ...?

  1. 222

  2. 224

  3. 291

  4. 298
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

nth term = a + (n – 1) d + ${\frac 12}$ (n – 1) (n – 2) c             = 3 + (n – 1) 3 + ${\frac 12}$(n2 – 3n + 2) 2             = 3 + 3n – 3 + n2 – 3n + 2             = n2 + 2 17th term = (17)2 + 2             = 289 + 2 = 291

Multiple choice median percentiles and quartiles range and mean deviation mode

If the 2nd term of a GM series is 25, the first and 3rd terms will be _____.

  1. 5,125

  2. 105,5

  3. 5,105

  4. 25,125
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Geometric mean between two numbers is the mean proportion of the series. 

Therefore, let the two numbers be a and b 
a/25 = 25/b 

=>ab= 25 x 25

Also , a/b = 1/25

 b = 25a

so, a (25a) = 25 x 25

=>a2 = 25

=> a = 5

Now, b = 25(5)

        b = 125  

Therefore, the first and the 3rd terms are 5 and 125 respectively. 

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Find the sum of the following geometric series:
$ \sqrt{7}, \sqrt{21}, 3\sqrt{7},...$ to n terms

  1. $ \sqrt{7}\left ( \dfrac{3^{-n/2}-1}{\sqrt{3}-1} \right )$
  2. $ \sqrt{6}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
  3. $ \sqrt{7}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
  4. $ \sqrt{5}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $ S _n$ denote the sum of n terms of the G.P. $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, ...,$ 


Cleraly, the given series is a $G.P.$ with first term$=a=\sqrt7$ and common ratio$=r=\sqrt 3$

Then,
$ S _n = \sqrt{7} \left { \dfrac{\left ( \sqrt{3} \right )^{n}-1}{\sqrt{3}-1} \right } = \sqrt{7} \left ( \dfrac{3^{n/2}-1}{3^{1/2}-1} \right )$

Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

Select the correct alternative from the given ones that will complete the series.
$0, 7, 26, 63, 124, ?$

  1. $251$
  2. $125$
  3. $215$
  4. $512$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\underset {(1^{3} - 1)}{0}\rightarrow \underset {(2^{3} - 1)}{7}\rightarrow \underset {(3^{3} - 1)}{26}\rightarrow \underset {(4^{3} - 1)}{63}\rightarrow \underset {(5^{3} - 1)}{124} \rightarrow \underset {(6^{3} - 1)}{215}$.