# Number Sequence Questions

Number series also are known as number sequence is the most important part of the reasoning ability section. Number series questions measure your ability to reason without words. At least 2 or 3 questions from number pattern series are asked in competitive exams like SSC, UPSC, CDS, bank exams such as IBPS, SBI PO, etc. If you master this topic, you can easily solve the letter or alphabet series questions. There are many ways to solve number sequence problems. Sometimes it will be very easy to solve but sometimes it will take a long time to identify patterns and solve these questions. The best way to master these types of questions is to practice many different types of number series questions so that you will be able to build different logic inside your mind while tackling these questions in the competitive exams.

In this section, we have tried to provide different types of number series questions with a lot of practice problems. We have also tried to provide you with some shortcuts, tricks, and notes that will help to revise this topic quickly before your exam. We have also included some mock test papers and a group of questions that you can solve in regular intervals to help you brush up critical thinking and reasoning skills.

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## What are Number Series or Number Sequences?

The number series are the problems which are based number pattern or sequence in the series which follow a logical pattern based on mathematical operations such as AP, GP, AGP, etc.

Examples:

1, 4, 9, 16, ?

3, 5, 8, 13, 21, ?

1, 2, 4, 7, ?

## Different types of number series questions and how to solve them?

### Type I – Arithmetic progression series

In this type of series, the difference between consecutive terms remains constant.

a, a+d, a+2d, …..

The nth term of the AP =a+ (n−1) d,

where a is the first term and d is the common difference.

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 3, 7, ? , 15, 19, ? , 27, 31
• a) 11, 22
• b) 11, 23,
• c) 12, 24
• d) 12, 23
• e) 12, 22

b) 11, 23

Show Solution

To solve this question, first we will take difference of consecutive terms

3, 7, ? , 15, 19, ? , 27, 31

Difference between consecutive terms become 4, ?, ?, 4, ? , 4

From above we can notice that the difference between the consecutive terms is 4.

Therefore this is arithmetic series with common difference of 4.

We know that the typical arithmetic series is

a, a+d, a+2d, a+3d, ….

In this question, a is 3 and the common difference is 4

So our series becomes

3, 3+4, 3+2*4, 3+3*4, 3+4*4, 3+5*4, 3+6*4, 3+7*4

3, 7, 11, 15, 19, 23, 27, 31

So our missing terms are 11 and 23.

Hence, option b) 11, 23 is correct.

#### Practice Problems

Question 1
1. Select the option that can replace the question marks (?) in the following series.
2. 3, 6, 9, 12, 15, ? .
3. a) 16
4. b) 18
5. c) 19
6. d) 21
1. b) 18

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 66, 59, 52, 45, 38, ? .
3. a) 31
4. b) 32
5. c) 35
6. d) 41
1. a) 31

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 567, 542, 517, 492, ? .
3. a) 467
4. b) 477
5. c) 483
6. d) 499
1. a) 467

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 95, 114, 133, 152, ? .
3. a) 162
4. b) 161
5. c) 171
6. d) 163
1. c) 171

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 149, 130, 111, 92, ? .
3. a) 71
4. b) 73
5. c) 72
6. d) 74
1. b) 73

### Type II – Geometric progression series

In this type of series, the ratio of consecutive terms remains constant.

a, ar, ar2, ar3, …..,

where, r is the common ratio.

Now, missing term (nth term) =arn−1

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 3, 9, 27, 81, ?
• a) 99
• b) 243
• c) 117
• d) 180

b) 243

Show Solution

To solve this question, first we will take ratio of consecutive terms

3, 9, 27, 81, ?

Ratio – 9/3=3, 27/9=3, 81/27=3, ?

From above we can notice that the ratio of the consecutive terms is 3.

Therefore this is a geometric series with common ratio of 3.

We know that the typical geometric series is

a, ar, ar2, ar3, …..,

In this question, a is 3 and the common ratio is 3

So our series becomes

3, 3*3, 3*32, 3*33, 3*34,

3, 9, 27, 81, 243

So our missing term is 243.

Hence, option b) 243 is correct.

#### Practice Problems

Question 1
1. Look at this series: 4, 36, 324, 2916, ? .
2. what number should come next?
3. a) 26244
4. b) 8748
5. c) 52488
6. d) 78732
1. a) 26244

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 1, 2, 4, 8, ?, 32, 64, ?, 256.
3. a) 24, 128
4. b) 16, 80
5. c) 24, 80
6. d) 16, 128
1. d) 16, 128

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 10, 30, 90, ? , ?, 2430
3. a) 180, 270
4. b) 270, 910
5. c) 270, 810
6. d) 270, 540
1. c) 270, 810

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 4, 2, 1, 0.5, 0.25, ? .
3. a) 0.125
4. b) 0.05
5. c) 0.625
6. d) 0.0625
1. a) 0.125

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 1, −3, 9, −27, ?, −243
3. a) 81
4. b) -81
5. c) 54
6. d) -54
1. a) 81

### Type III – Harmonic progression series

This series is formed by the reciprocal of arithmetic series in which reciprocal of consecutive terms have a common difference.

1/a, 1/(a+d), 1/(a+ 2d), 1/(a+ 3d), …..,

where a is the first term and d is the common difference

Now, missing term (nth term) = 1/(a+(n-1)d)

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 1, 1/2, 1/3, 1/4, 1/5, ?
• a) 1/6
• b) 1/7
• c) 1/8
• d) 1/9

a) 1/6

Show Solution

To solve this question, first reciprocate all the terms

1, 1/2, 1/3, 1/4, 1/5, ? becomes 1, 2, 3, 4, 5, ?

To solve this question, now we will take difference of consecutive terms

1, 2, 3, 4, 5, ?

difference of consecutive terms – 1, 1, 1, 1, ?

From above we can notice that the difference between the consecutive terms is 1.

Therefore this is arithmetic series with common difference of 1.

We know that the typical arithmetic series is

a, a+d, a+2d, a+3d, ….

In this question, a is 1 and the common difference is 1

So it is confirmed that the original series is reciprocal of above arithmetic series and is harmonic series.

Therefore, our missing term can be found using

1/(a+(n-1)d)

Where, n is our missing term number (in this case, it is 6)

a = 1, d = 1

1/(a+nd) => 1/(1+(6-1)*1) => 1/6

So our series becomes

1, 1/2, 1/3, 1/4, 1/5, 1/6

Hence, option a) 1/6 is correct.

#### Practice Problems

Question 1
1. Look at this series:
2. 12, 8, 4, 3, 12/5, ? .
3. what number should come next?
4. a) 12/7
5. b) 2
6. c) 3
7. d) 4
1. b) 2

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 1/3, 1/7, 1/11, ?,
3. a) 1/10
4. b) 1/11
5. c) 1/13
6. d) 1/14
1. c) 1/14

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 3, 1, 1/3, 1/9, ?
3. a) 1/8
4. b) 1/10
5. c) 1/27
6. d) 1/81
1. c) 1/27

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 60, 30, 20, 15, 12, ? .
3. a) 8
4. b) 9
5. c) 10
6. d) 11
1. c) 10

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 1/149, 1/130, 1/111, 1/92, ?
3. a) 1/71
4. b) 1/73
5. c) 1/72
6. d) 1/74
1. b) 1/73

### Type IV – Product and Division Series

In this type of series, the terms are formed by the product of numbers in particular order.

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 3, 8, 15, 24, ?
• a) 27
• b) 30
• c) 33
• d) 35
• e) 39

d) 35

Show Solution

To solve this question, first we will write the series in their multiples as follows:

3×1, 4×2, 5×3, 6×4, ?

From above we can notice that there are two series which are multiplied and merged together

Series I x Series II

(3, 4, 5, 6 ,? ) x (1, 2, 3, 4, ?)

Now let’s solve these series separately

Series I : 3, 4, 5, 6, ?

This is arithmetic progression with common difference of 1.

So the series becomes 3, 4, 5, 6, 7

Series II: 1, 2, 3, 4, ?

This is also a arithmetic progression with common difference of 1.

So the series II becomes 1, 2, 3, 4, 5

Now merge these series together

(3, 4, 5, 6, 7 ) x (1, 2, 3, 4, 5)

So our series becomes

3×1, 4×2, 5×3, 6×4, 7×5

3, 8, 15, 24, 35

So our missing term is 35.

Hence, option d) 35 is correct.

Question 1
1. Look at this series: 3, 6, 9, 12, 15, ? .
2. what number should come next?
3. a) 16
4. b) 18
5. c) 19
6. d) 21
1. b) 18

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 3, 6, 9, 15, 21, ?
3. a) 24
4. b) 33
5. c) 30
6. d) 36

1. b) 33

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 16, 24, 32, 40, 36, 48, 56, 60, 64, ? .
3. a) 68
4. b) 72
5. c) 76
6. d) 80
1. b) 72

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 3, 12, 27, 48, ? .
3. a) 60
4. b) 66
5. c) 75
6. d) 72
1. c) 75

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 2, 16, 54, 128, ? .
3. a) 182
4. b) 250
5. c) 150
6. d) 172
1. b) 250

### Type V: Arithmetico–geometric sequence

This series is consists of product of arithmetic progression and geometric progression.

a, (a+d)r, (a+2d)r2, (a+3d)r3,…,[a+(n−1)d]rn−1

Where, a is the initial term,

d is the common difference, and r is the common ratio

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 3, 36, 189, 810, ?
• a) 3159
• b) 2430
• c) 7290
• d) 5670

a) 3159

Show Solution

To solve this question, first we will write the series in their multiples as follows:

1×3, 4×32, 7×33, 10×34, ?

From above we can notice that there are two series which are multiplied and merged together

Series I x Series II

(1, 4, 7, 10,? ) x (3, 32, 33, 34, ?)

Now let’s solve these series separately

Series I : 1, 4, 7, 10, ?

This is arithmetic progression with common difference of 3.

So the series becomes 1, 4, 7, 10, 13

Series II: 3, 32, 33, 34, ?

This is also a geometric progression with common ratio of 3.

So the series II becomes 3, 32, 33, 34, 35

Now merge these series together

(1, 4, 7, 10, 13) x (3, 32, 33, 34, 35)

So our series becomes

1×3, 4×32, 7×33, 10×34, 13×35

3, 36, 189, 810, 3159

So our missing term is 3159.

Hence, option a) 3159 is correct

#### Practice Problems

Question 1
1. Look at this series:
2. 2, 8, 24, 64, ?.
3. what number should come next?
4. a) 160
5. b) 150
6. c) 180
7. d) 200

1. a) 160

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 15, 63, 243, ?, 3159
3. a) 810
4. b) 891
5. c) 890
6. d) 801
1. b) 891

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 4, 24, 80, 224, ? .
3. a) 448
4. b) 372
5. c) 576
6. d) 628
1. c) 576

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 5, 50, 375, 2500, ?.
3. a) 25625
4. b) 31525
5. c) 15625
6. d) 16625
1. c) 15625

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 4, 20, 64, 176, ? .
3. a) 452
4. b) 448
5. c) 488
6. d) 542
1. b) 448

### Type VI: Perfect square Series

This series are consists of terms who are perfect squares or perfect square along with mathematical operations such as addition and subtraction.

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 1, 4, 9, 16, 25, 36, ?
• a) 39
• b) 49
• c) 59
• d) 64

b) 49

Show Solution

To solve this question, first we will write the series in their multiples as follows:

1×1, 2×2, 3×3, 4×4, 5×5, ?

12, 22, 32, 42, 52, ?

From above we can notice that it is series with perfect square series.

X12, X22, X32, X42, X52, ?

Where X1, X2, X3, X4, X5 forms arithmetic progression with common difference of 1.

So the missing term in above arithmetic progression series becomes X6 = 7

And missing term in our original series will be perfect square of X6.

The missing term becomes X62 = 72 = 49.

Our series becomes 1, 4, 9, 16, 25, 36, 49.

So our missing term is 49.

Hence, option b) 49 is correct.

#### Practice Problems

Question 1
1. Select the number that can replace the question mark (?) in the following series.
2. 169, 121, 49, ?, 9, 4.
3. a) 25
4. b) 36
5. c) 64
6. d) 16
1. a) 25

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 16, 36, 64, 81, 100, 144, ?
3. a) 159
4. b) 196
5. c) 169
6. d) 225
1. b) 196

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 16, 36, 64, ?, 144, 196, 225.
3. a) 121
4. b) 81
5. c) 100
6. d) 120
1. c) 100

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 3, 6, 11, 18, 27, 38, ?.
3. a) 66
4. b) 50
5. c) 51
6. d) 64
1. c) 51

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 0, 3, 8, 15, 24, ?, 48.
3. a) 36
4. b) 35
5. c) 34
6. d) 32
1. b) 35

### Type VII: Perfect Cube Series

This series are consists of terms which are formed by perfect cube along with mathematical operations such as addition and subtraction.

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 1, 8, 27, 64, 125, 216, ?
• a) 249
• b) 343
• c) 349
• d) 464

b) 343

Show Solution

To solve this question, first we will write the series in their multiples as follows:

1x1x1, 2x2x2, 3x3x3, 4x4x4, 5x5x5, ?

13, 23, 33, 43, 53, ?

From above we can notice that it is series with perfect cube series.

X13, X23, X33, X43, X53, ?

Where X1, X2, X3, X4, X5 forms arithmetic progression with common difference of 1.

So the missing term in above arithmetic progression series becomes X6 = 7

And missing term in our original series will be perfect cube of X6.

The missing term becomes X63 = 73 = 343.

Our series becomes 1, 8, 27, 64, 125, 216, 343.

So our missing term is 343.

Hence, option b) 343 is correct.

#### Practice Problems

Question 1
1. Select the number that can replace the question mark (?) in the following series.
2. 2197, 1331, 343, ?, 27, 8.
3. a) 25
4. b) 125
5. c) 225
6. d) 75
1. b) 125

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 64, 216, 512, 729, 1000, 1728, ?
3. a) 2700
4. b) 2744
5. c) 2796
6. d) 2728
1. b) 2744

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 64, 216, 512, ?, 1728, 2744, 3375.
3. a) 1211
4. b) 810
5. c) 1000
6. d) 1200

1. c) 1000

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 2, 9, 28, 65, 126, 217, ?.
3. a) 225
4. b) 256
5. c) 343
6. d) 379
1. c) 343

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 342, ?, 124, 63, 26, 7, 0
3. a) 216
4. b) 215
5. c) 256
6. d) 169
1. b) 215

### Type VIII: Square difference number series

In these type of series, the difference between the consecutive terms forms a perfect square series or a series having squares along with mathematical operations such as addition and subtraction.

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 5, 9, 18, 34, 49, ?
• a) 64
• b) 56
• c) 85
• d) 81

c) 85

Show Solution

To solve this question, first we find out the difference between the consecutive terms

Difference – 4, 9, 25, ?

From above we can notice that the difference between the consecutive terms forms a perfect square series.

4, 9, 16, 25, ?

So the missing term in above perfect square series will become 36.

Now we know that the difference the missing term i.e, last term and second last term is 36.

Missing term = second last term + difference between last and second last term

Missing term = 49+36 => 85

Our series becomes 5, 9, 18, 34, 49, 85.

So our missing term is 85.

Hence, option c) 85 is correct.

#### Practice Problems

Question 1
1. Select the number that can replace the question mark (?) in the following series.
2. 64, 39, 23, 14, ?.
3. a) 8
4. b) 9
5. c) 10
6. d) 4
1. c) 10

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 391, 291, 210, 146, 97, ?
3. a) 60
4. b) 61
5. c) 63
6. d) 65
1. b) 61

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 7, 16, 32, ?, 83, 132.
3. a) 49
4. b) 47
5. c) 44
6. d) 42

1. b) 47

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 20, 24, 33, 49, ?.
3. a) 74
4. b) 64
5. c) 81
6. d) 89
1. a) 74

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 213, 149, ?, 64, 39, 23
3. a) 121
4. b) 81
5. c) 120
6. d) 100
1. d) 100

### Type IX: Cube difference number series

In these type of series, the difference between the consecutive terms forms a perfect cube series or a series having cubes along with mathematical operations such as addition and subtraction.

Solved Question
• Select the number that can replace the question mark (?) in the following series.
• 7, 15, 42, 106, 231, ?
• a) 289
• b) 449
• c) 447
• d) 441

c) 447

Show Solution

To solve this question, first we find out the difference between the consecutive terms

Difference – 8, 27, 64, 125, ?

From above we can notice that the difference between the consecutive terms forms a perfect cube series.

8, 27, 64, 125, ?

23, 33, 43, 53, ?

So the missing term in above perfect square series will become 63 = 216.

Now we know that the difference the missing term i.e, last term and second last term is 216.

Missing term = second last term + difference between last and second last term

Missing term = 231+216 => 447

Our series becomes 7, 15, 42, 106, 231, 447.

So our missing term is 447.

Hence, option c) 447 is correct.

#### Practice Problems

Question 1
1. Select the number that can replace the question mark (?) in the following series.
2. 2, 10, 37, 101, ?, 569.
3. a) 226
4. b) 125
5. c) 144
6. d) 169
1. a) 226

Question 2
1. Select the number that can replace the question mark (?) in the following series.
2. 289, 164, 100, 36, ?
3. a) 10
4. b) 9
5. c) 16
6. d) 8

1. b) 9

Question 3
1. Select the number that can replace the question mark (?) in the following series.
2. 1, 3, 12, 40, 105, 231, ?.
3. a) 244
4. b) 429
5. c) 458
6. d) 369
1. c) 458

Question 4
1. Select the number that can replace the question mark (?) in the following series.
2. 0, 7, 33, 96, 220, 435, ?.
3. a) 444
4. b) 555
5. c) 666
6. d) 777
1. d) 777

Question 5
1. Select the number that can replace the question mark (?) in the following series.
2. 1, 28, 153, 496, 1225, ?.
3. a) 1331
4. b) 1728
5. c) 2556
6. d) 1302
1. c) 2556

## Number Series Tricks, Shortcuts and Notes to Solve Alphabet Series Questions

1. Do not spend more than one minute on these questions. These are those types of questions in which it is possible that you can identify the pattern very easily or sometimes it might take a long time. Therefore always spend fix amount of time on these questions
2. Only more and more practice will help you to master these questions
3. While solving look if any of the following important series are present
• AP series
• GP series
• AGP series
• Harmonic series
• Product series
• Fibonacci Sequence – 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
• Triangular number sequence – 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, …
• Prime number sequence – 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, …
• Composite number Sequence – 4, 6, 8, 9, 10, 12, 14, 15, 16, …
• Factorial series – 1, 2, 6, 24, 120, …
• Perfect Square Series – 1, 4, 9, 16, 25, 36, …
• Perfect Cube Series – 1, 8, 27, 64, 125, …
4. Try to observe if there are any two alternate series are present
5. Sometimes it is also possible that there will two-stage series present. In these cases, the difference or ratio of the consecutive numbers create another number series
6. Look for series having a mix of Alphabet and number series. These are generally easy to solve. In these questions look for a relation between the number and alphabet.
7. If the series is increasing or decreasing exponentially then, look for product series, perfect square, perfect cube, AGP, or GP series.
8. If the series is not increasing or decreasing exponentially then, look for AP series or difference between consecutive numbers.