Test: Number Series - SSC CGL MCQ

1st term = 1² = 1
2nd term = 2² = 4
3rd term = 3² = 9
4th term = 4² = 16
5th term = 5² = 25
Thus, Missing number = 6² = 36

1st term = 20 - 0 = 20
2nd term = 20 - 1 = 19
3rd term = 19 - 2 = 17
4th term = 17 - 3 = 14
5th term = 14 - 4 = 10
6th term = 10 - 5 = 5

Thus, Missing number is 4th term = 14

Clearly, the given series consists of prime numbers starting from 2. The prime number after 11 is 13.

Thus, the missing number is 13.

1st term = 6
2nd term = 6 + 5 = 11
3rd term = 11 + 10 = 21
4th term = 21 + 15 = 36
5th term = 36 + 20 = 56

Thus, Missing Number is 6th term = 56 + 25 = 81

1st term = 3 * 1 = 3
2nd term = 3 * 3 = 9
3rd term = 3 * 9 = 27
4th term = 3 * 27 = 81

Thus, Missing number is 5th term = 3 * 81 = 243

1st term = 2
2nd term = 2 + 3 = 5
3rd term = 5 + 4 = 9
4th term = 9 + 5 = 14
5th term = 14 + 6 = 20
6th term = 20 + 7 = 27

Thus, Missing number is 4th term = 14

1st term = 5
2nd term = 5 + (4*1) = 5 + 4 = 9
3rd term = 9 + (4*2) = 9 + 8 = 17
4th term = 17 + (4*3) = 17 + 12 = 29
5th term = 29 + (4*4) = 29 + 16 = 45

Thus, the missing number is 6th term = 45 + (4*5) = 45 + 20 = 65

1st term = 1
2nd term = 1 + 5 = 6
3rd term = 6 + (5+4) = 6 + 9 = 15
4th term = 15 + (9+4) = 15 + 13 = 28
5th term = 28 + (13+4) = 28 + 17 = 45
6th term = 45 + (17+4) = 45 + 21 = 66
7th term = 66 + (21+4) = 66 + 25 = 91

Thus, the missing number is 4th term= 15 + (9+4) = 15+13 = 28

1st term = 0.5
2nd term = 0.5 * 3 = 1.5
3rd term = 1.5 * 3 = 4.5
4th term = 4.5 * 3 = 13.5

Thus, the missing number is 5th term = 13.5 * 3 = 40.5

The sequence is a combination of two series:

I. 19, 38, 114, (____)
II. 2, 3, 4

Considering Series I:
1st term = 19
2nd term = 19*2 = 38
3rd term = 38*3 = 114
Thus, the missing number is 4th term = 114 * 4 = 456

1st term = 1
2nd term = 1 + (1) = 2
3rd term = 2 + (1) = 3
4th term = 3 + (3) = 6
5th term = 6 + (3) = 9
6th term = 9 + (9) = 18
7th term = 18 + (9) = 27
8th term = 27 + (27) = 54

Thus, the missing term is 7th term = 27

1st term = 12
2nd term = 12 + 20 = 32
3rd term = 32 + 40 = 72
4th term = 72 + 80 = 152

Thus, the missing term is 5th term = 152 + 160 = 312

The sequence is a combination of two series:
I. 3, 5, 7, 9
II. 6, 20, 42, (____)

Consider II sequence
1st term = 6
2nd term = 6 + 14 = 20
3rd term = 20 + 14 = 34

Thus, the missing term is 4th term = 34 + 14 = 48

1st term = 1
2nd term = 3
3rd term = 4
4th term = 1 + 3 + 4 = 8
5th term = 3 + 4 + 8 = 15
6th term = 4 + 8 + 15 = 27

Thus, the missing term is 7th term = 8 + 15 + 27 = 50

1st term = 95
2nd term = 95 + 20.5 = 115.5
3rd term = 115.5 + 22.5 = 138
4th term = 138 + 24.5 = 162.5
5th term = 162.5 + 26.5 = 189

Thus, the missing term is 4th term = 162.5

1st term = 24
2nd term = 24 + [6 x (6 + 0)] = 60
3rd term = 60 + [6 x (6 + 4)] = 120
4th term = 120 + [6 x (6 + 9)] = 210

Thus, the missing term is 5th term = 210 + [6 x (6 + 15)] = 210 + 126 = 336

1st term = 34/2 +1 = 18
2nd term = 18/2 +1 = 10
3rd term = 10/2 +1 = 6
4th term = 6/2 + 1  = 4

Thus, the missing number is 5th term = 4/2 + 1 = 3

1st term = 0.5
2nd term = 0.5 + 0.05 = 0.55
3rd term = 0.55 + 0.1 = 0.65
4th term = 0.65 + 0.15 = 0.8

Thus, missing number is 5th term = 0.8 + 0.2 = 1

The given series consists of consecutive prime numbers starting from 17. The next prime number after 29 is 31.

Thus, the missing number is 31.

1st term = 3
2nd term = 3 + 9 = 12
3rd term = 12 + 15 = 27
4th term = 27 + 21 = 48
5th term = 48 + 27 = 75
6th term = 75 + 33 = 108

Thus, the missing term is 7th term = 103 + 39 = 147