# Number Series Questions for RRB NTPC Set-2 PDF

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## Number Series Questions for RRB NTPC Set-2 PDF

Download RRB NTPC Number Series Questions and Answers set – 2 PDF. Top 15 RRB NTPC Number series questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: Find the next number in the series:
12, 36, 108, 324, 972, ___

a) 2712

b) 2916

c) 3016

d) 3200

Question 2: Find the missing number in the series:
4, 5, 9, ?, 23, 37, 60, 97

a) 11

b) 13

c) 19

d) 14

Question 3: Find the next number in the series:
5, 16, 27, 38, 49, 60, ____

a) 70

b) 71

c) 85

d) 82

Question 4: What comes next in the given series:2,3,5,7,11,13,17

a) 20

b) 19

c) 21

d) 22

Question 5: What comes next in the given series:3,8,15,24,35,48,

a) 63

b) 80

c) 56

d) 60

Question 6: What comes next in the given series:3,8,15,24,35,48,63,80

a) 90

b) 91

c) 99

d) 97

Question 7: What comes next in the given series:1,2,4,7,11,16,22,29,37,46,

a) 55

b) 56

c) 52

d) 59

Question 8: Complete the following series: 11, 15, 20, 26, 33, ?

a) 45

b) 43

c) 41

d) 40

Question 9: Complete the following series: 10, 17, 25, 34, 44, ?

a) 54

b) 55

c) 64

d) 61

Question 10: Find the missing term in the series:
981, 1089, 1225, 1369, ?

a) 1461

b) 1521

c) 1681

d) 1491

Question 11: Find the missing term in the series:
3, 7, 13, 21, ?

a) 31

b) 30

c) 29

d) 28

Question 12: Complete the following series:
19, 79, 319, 1279, ?

a) 5009

b) 5109

c) 5111

d) 5119

Question 13: Complete the following series:
5, 11, 8, 13, 11, ?

a) 15

b) 5

c) 10

d) 8

Question 14: Complete the following series:
50, 255, 1280, 6405, ?

a) 30100

b) 24590

c) 30300

d) 32030

Question 15: Complete the following series:
11, 18, 28, 41, 57, ?

a) 76

b) 74

c) 86

d) 84

The differences between consecutive terms are 24, 72, 216, 648. Hence, the differences are increasing in an exponential manner. Thus, the series must be a Geometric Progression. a=12 and r=36/12 = 3.

Thus, the terms of the series are 12, 12*3, $12*3^2$, $12*3^3$, $12*3^4$. Thus, the next term of the series = 972 * 3 = 2916.

The differences between consecutive terms are 1, 4 and 14, 23 and 37. We see that terms of the series are appearing in the list of differences. Hence, the series is formed by adding previous terms of the series.
i.e. 4+5=9, 5+9=14, 9+14=23, 14+23=37, 23+37=60 and 37+60=97.

Hence, the missing number is 14.

The differences between consecutive terms is 16-5=11, 27-16=11, 38-27-11, 49-38=11 and 60-49=11. Hence, the difference is a constant. Thus, the series is an Arithmetic Progression.

The next number in the series would be 60+11=71.

The given series is the list of prime numbers. The next prime number after 17 is 19.

The given series is of the form n(n+2) = 1*3, 2*4, 3*5, 4*6 and so on.
The next term would be 7*9 = 63

The given terms are of the format $n^2-1$.
3 =$2^2-1$
8 =$3^2-1$
15 =$4^2-1$
24 =$5^2-1$
35 =$6^2-1$
and so on
The next term would be $10^2-1 = 99$

The difference of two consecutive terms are in AP.
2-1 = 1, 4-2 = 2, 7-4 = 3, 11-7 = 4, 16-11 = 5, 22-16 = 6, 29-22 = 7, 37-29 = 8, 46-37 = 9
x-46 = 10
x = 56

11+4 = 15 ; 15+5 = 20 ; 20+6 = 26 ; 26+7 = 33; 33+8 = 41

10+7 = 17; 17+8 = 25 ; 25+9 = 34 ; 34+10 = 44 ; 44+11 = 55

981 = $31^2$
1089 = $33^2$
1225 = $35^2$
1369 = $37^2$
1521 = $39^2$

7 – 3 = 4
13 – 7 = 6
21 – 13 = 8
x – 21 = 10
=> x = 31

19*4+3 = 79 ; 79*4+3 = 319; 319*4+4 = 1279 ; 1279*4+3 = 5119

First, third and 5th letter are forming an A.P. with the common difference of 3
Second, third and sixth letter are also forming an A.P. with the common difference of 2