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# Most Expected Number Series Questions for IBPS RRB Clerk PDF

Download Top-15 Banking Exams Number Series Questions PDF. Banking Exams Number Series questions based on asked questions in previous exam papers very important for the Banking  exams.

Instructions

Find the wrong number in given series sequence.

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: Which number comes in the place of the question mark (?)16, 25, 36, 49, 64, ?

a) 121

b) 100

c) 81

d) 144

e) 196

Question 7: Which number comes in the place of the question mark (?)1, 27, 125, 343, ?

a) 512

b) 1000

c) 729

d) 216

e) 1024

Question 8: Which number comes in the place of the question mark (?)11, 44, 99, 176, ?

a) 274

b) 266

c) 275

d) 265

e) 300

Question 9: Which number comes in the place of the question mark (?)5, 25, 61, 113, ?

a) 144

b) 181

c) 165

d) 200

e) 245

Question 10: Which number comes in the place of the question mark (?)2, 6, 12, 20, 30, ?

a) 50

b) 44

c) 45

d) 42

e) 55

Instructions

What should come in place of the question mark (?) in the following number series ?

Question 11: 353 354 351 356 349 ?

a) 348

b) 358

c) 338

d) 385

e) 340

Question 12: 1 5 13 29 ? 125 253

a) 83

b) 69

c) 61

d) 65

e) 81

Question 13: 45 57 81 117 165 ?

a) 235

b) 215

c) 205

d) 245

e) 225

Question 14: 17 18 26 53 117 ? 458

a) 342

b) 142

c) 257

d) 262

e) 242

Question 15: $\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, 1\frac{1}{4}, 1\frac{1}{2}, 1 \frac{3}{4}, ?$

a) $2$

b) $4$

c) $6$

d) $1\frac{1}{5}$

e) $1\frac{2}{3}$

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.

6+6 = 12
12+9 = 21
21+12 = 33
33+15 = 48
48+18 = 66.
Hence, 49 is wrong number and it should be 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
All the numbers except 214 are following a pattern of $n\times (n+1) + (n+2)$. Hence, 214 is wrong number.

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

$4^2 = 16$ ; $5^2=25$ ; $6^2 = 36$ ; $7^2=49$ ; $8^2 = 64$ ; $9^2 = 81$

$1^3 = 1$ ; $3^3 = 27$ ; $5^3 = 125$ ; $7^3=343$ ; $9^3 = 729$

$1 \times11 = 11$ ; $2 \times 22=44$ ; $3 \times 33=99$ ; $4 \times 44=176$ ; $5 \times 55=275$

$1^2 + 2^2 = 5$; $3^2 + 4^2 = 25$ ; $5^2 +6^2 = 61$ , $7^2 + 8^2 = 113$ ; $9^2 + 10^2 = 181$

2+4=6 ; 6+6 = 12 ; 12+8 = 20 ; 20+10=30 ; 30+12 = 42

There are two arithmetic progression series,

With 353,351,349… having a difference of -2

and 354, 356… having a difference of 2

Hence, the required term is 358

The generalized formula for nth term is, previous term + 2^n

Here the required term is 5th in the series.

Hence, 29 + 2^5 = 61

For the series the nth term is ‘last term + 12*(n-1)’

In the series, we are required to find, 6th term.

Hence, 6th term is 165+12*(5) = 225.

Here, each sucessive nth term is ‘previous term + (n-1)³’

We are supposed to find 6th term.

Hence, 6th term = 117 + (6-1)^3 = 117 + 125 = 242

The given series is Arithmetic Progression with common difference $\frac{1}{4}$.
Hence, the next term will be $1\frac{3}{4}$ + $\frac{1}{4}$ = 2