# IBPS Clerk Number Series Questions PDF

0
10548

Number series questions are very important for IBPS exams like IBPS Clerk and PO. A series of numbers will be given and you will asked to find out the next or missing or wrong number. Each series is formed from a particular progression/logic. Here comes a list of questions in which one number is wrong and rest of the terms follow the series.

IBPS Clerk Number Series Questions PDF:

You can download the Number Series questions PDF or you can go through the details below.

Take Free IBPS Clerk Mock Test

Go to Free Banking Study Material (15,000 Solved Questions)

Instructions:

In each of these question a number series is given.In each series only one number is wrong.Find out the wrong numbers

Question 1:

1, 4, 15, 64 325, 1955

A. 15
B. 64
C. 325
D. 1955
E. None of these

Question 2:

6, 12, 21, 33, 49, 66

A. 21
B. 33
C. 49
D. 66
E. None of these

Question 3:

6, 11.5, 19, 28.5, 41

A. 6
B. 11.5
C. 41
D. 28.5
E. None of these

Question 4:

5, 26, 82, 214, 401, 702

A. 26
B. 82
C. 214
D. 401
E. None of these

Question 5:

5,20,73,274,1049

A. 20
B. 73
C. 274
D. 1049
E. None of these

Question 6:

3601, 3602, 1803, 604, 154, 36, 12

A. 3602
B. 1803
C. 604
D. 154
E. 36

Question 7:

4, 12, 42, 196, 1005, 6066, 42511

A. 12
B. 42
C. 1005
D. 196
E. 6066

Question 8:

2, 8, 12, 20, 30, 42, 56

A. 8
B. 42
C. 30
D. 20
E. 12

Question 9:

32, 16, 24, 65, 210, 945, 5197.5

A. 945
B. 16
C. 24
D. 210
E. 65

Question 10:

7, 13, 25, 49, 97, 194, 385

A. 13
B. 49
C. 97
D. 194
E. 25

Solutions to Number Series Questions for IBPS Clerk:

4 = 1*2+2 ; 15 = 4*3+3 ; 64 = 15*4+4 ; 325 = 64*5+5 ; 1956 = 325*6+6
The nth term is of the form,
$T_n=(T_{n-1}\times n)+n$
The last term does not follow the pattern and is thus the wrong number in the sequence.

2*2+2 = 6 ; 3*3+3 = 12 ; 4*4 + 5= 21 ; 5*5 +8 = 33 ; 6*6 + 12 = 48

1.5*2 + 3 = 6
2.5*3 + 4 = 11.5
3.5*4 + 5 = 19
4.5*5 + 6 = 28.5
5.5*6 + 7 = 40

1*2+3=5 ; 4*5+6 = 26 ; 8*9+10 = 82 ; 13*14+15 = 212 ; 19*20 + 21 =401 ; 26*27 +28 = 702

1^2 + 4= 5 ; 2^2 + 16 = 20 ; 3^2 + 64 = 73 ; 4^2 + 256 = 272 ; 5^2 + 1024 = 1049

In the given series, nth term is obtained by – (Previous term/(n-1) ) +(n-1).
Hence the 5th term should be (604/4)+4 = 155.
However 5th term is given as 154.
Hence, 154 is wrong term in the given series.

In this question, the nth term is obtained by (previous term*n + n^2).
Hence, after 12, the number would be (12*3+9)= 45
Therefore, the correct option is 45.

Here the consecutive difference between the terms is incremented by 2 for the series.
Hence, after 2, the number should be 6.

Here, in this series, the nth term is (previous term + $6*2^{n-2}$)
Hence, in place of 194 the term should be $(97+6*2^{4}) = 193$