# (Study Materials) Mental Ability/Reasoning (Verbal) : Series Completion

## (Study Materials) Mental Ability/Reasoning (Verbal) : Series Completion

## Number Series

**Prime Number Series:**

**Example 1. 4, 9, 25, 49, 121, 169,…
**(a) 324 (b) 289

(c) 225 (d) 196

Solution. (b) The given series is a consecutive square of prime number series. The next prime number is 289.

**Example 2. 5, 7, 13, 23, …
**(a) 25 (b) 27

(c) 29 (d) 41

Solution. (d) The difference between prime numbers is increasing. 7 is next prime to 5; 13 is second to next prime to 7; 23 is third to next to 13. Hence, next should be fourth to next prime to 23. Hence, required number is 41.

**Multiplication Series:**

**Example 3. 4, 8, 16, 32, 64… 256
**(a) 96 (b) 98

(c) 86 (d) 106

Solution. (a) The numbers are multiplied by 2 to get the next number.

64 × 2 = 128

**Example 4. 5, 20, 80, 320, … 1280
**(a) 5120 (b) 5220

(c) 4860 (d) 3642

Solution. (a) The numbers are multiplied by 4 to get the next number.

1280 × 4 = 5120

**Difference Series:**

**Example 5. 3,6,9,12,15,…. 21
**(a) 16 (b) 17

(c) 20 (d) 18

Solution. (d) The difference between the numbers is 3.

15 + 3 = 18

**Example 6. 55, 50, 45, 40,….30
**(a) 33 (b) 34

(c) 35 (d) 36

Solution. (c) The difference between the numbers is -5.

40 – 5 = 35

**Division Series:**

**Example 7. 5040, 720, 120, 24, ….2,1
**(a) 8 (b) 7

(c) 6 (d) 5

Solution. (c)

**Example 8. 16, 24, 36,… 81
**(a) 52 (b) 54

(c) 56 (d) 58

Solution. (b) Previous number × = Next number

**n2 Series**

Example 9. 4, 16, 36, 64, …. 144

(a) 112 (b) 78

(c) 100 (d) 81

Solution. (c) The series is square of consecutive even numbers. 22, 42,62, 82

Next number is 102 = 100

**Example 10. 1, 4, 9, 16, 25, 36, 49, … 81
**(a) 100 (b) 121

(c) 64 (d) 144

Solution. (c) The series is 12, 22, 32, 42, 52,62, 72,….

The next number is 82 = 64

**(n2 + 1) Series****Example 11. 17, 26, 37, 50, 65,….101
**(a) 82 (b) 75

(c) 78 (d) 90

Solution. (a) The series is 42 + 1, 52 +1, 62 + 1, 72 + 1, 82 + 1.

The next number is 92 + 1 = 82

**Example 12. 101, 401, 901, 1601, 2501, …. 4901
**(a) 2201 (b) 3301

(c) 4401 (d) 3601

Solution. (d) The series is 102 + 1, 202 +1, 302 + 1, 402 + 1, 502 + 1, etc.

The next number is 602 + 1 = 3601

**(n2 -1) Series**

**Example 13. 3, 8, 15, 24,…48
**(a) 32 (b) 33

(c) 34 (d) 35

Solution. (d) The series is 22 – 1, 32 –1, 42 – 1,52 – 1. etc.

The next number is 62 – 1 =35

**Example 14. 99, 80, 63,….35
**(a) 48 (b) 84

(c) 46 (d) 64

Solution. (a) The series is 102 -1, 92 -1, 82 -1, etc.

The next number is 72 – 1 = 48

*(n2 + n) Series*

**Example 15. 2, 6, 12, 20, 30,…. 56
**(a) 32 (b) 34

(c) 42 (d) 24

Solution. (c) The series is 12 + 1, 22 + 2, 32 + 3, 42 + 4, 52 + 5, etc.

The next number is 62 + 6 = 42

**Example 16. 110, 132, 156, 182,….
**(a) 212 (b) 201

(c) 211 (d) 210

Solution. (d) The series is 102 + 10, 112 + 11, 122 + 12, etc.

The next number is 142 + 14 = 210

**(n2 – n) Series**

**Example 17. 0, 2, 6, 12, 20,….42
**(a) 25 (b) 30

(c) 32 (d) 40

Solution. (b) The series is 12 – 1 = 0, 22 – 2 = 2, 32 – 3 = 6, etc.

The next number is 62 – 6 = 30

**Example 18. 90, 380, 870, 1560,…..
**(a) 2405 (b) 2450

(a) 2400 (d) 2455

Solution. (b) The series is 102 – 10, 202 – 20, 302 – 30, etc.

The next number is 502 – 50 = 2450

**n3 Series**

**Example 19. 1, 8, 27, 64,…. 216
**(a) 125 (b) 512

(c) 215 (d) 122

Solution. (a) The series is 13, 23, 33 , 43, etc.

The next number is 53 = 125

**Example 20. 1000, 8000, 27000, 64000,….
**(a) 21600 (b) 125000

(c) 152000 (d) 261000

Solution. (b) The series is 103 , 203, 303, 403, etc.

The next number is 503 = 125000

**(n3 + 1) Series**

Example 21. 2, 9, 28, 65,…217

(a) 123 (b) 124

(c) 125 (d) 126

Solution. (d) The series is 13 +1, 23 + 1, 33 + 1, etc.

The next number is 53 + 1 = 126

**Example 22. 1001, 8001, 27001, 64001, 125001,….
**(a) 261001 (b) 216001

(c) 200116 (d) 210016

Solution. (b) The series is 103 + 1, 203 + 1, 303 + 1, etc.

The next number is 603 + 1 = 216001

**(n3 -1) Series**

Example 23. 0, 7, 26, 63, 124,…

(a) 251 (b) 125

(c) 215 (d) 512

Solution. (c) The series is 13 – 1, 23 – 1, 33 – 1, etc.

The next number is 63 – 1 = 215

**Example 24. 999, 7999, 26999, 63999,….
**(a) 199924 (b) 124999

(c) 129994 (d) 999124

Solution. (b) The series is 103 – 1, 203 – 1, 303 – 1, etc.

The next number is 503 – 1 = 124999

**(n3 + n) Series**

**Example 25. 2, 10, 30, 68,….222
**(a) 130 (b) 120

(c) 110 (d) 100

Solution. (a) The series is 13 + 1, 23 + 2, 33 + 3, etc.

The next number is 53 + 5 = 130

**Example 26. 1010, 8020, 27030, 64040,….
**(a) 125500 (b) 125050

(c) 100255 (d) 120055

Solution. (b) The series is 103 + 10 = 1010, 203 + 20 = 8020, etc.

The next number is 503 + 50 = 125050

**(n3 – n) Series**

**Example 27. 0, 6, 24, 60,…. 210
**(a) 012 (b) 210

(c) 201 (d) 120

Solution. (d) The series is 13 – 1 = 0, 23 – 2 = 6, 33 – 3 = 24, etc.

The next number is 53 – 5 = 120

**Example 28. 990, 7980, 26970, 63960,….
**(a) 124500 (b) 124005

(c) 120045 (d) 124950

Solution. (d) The series is 103 – 10, 203 – 20, 303 – 30 etc.

The next number is 503 – 50 = 124950

## Letter Series - Type 1

One Letter Series Such series consists of one letter in each term and this series is based on increasing or decreasing positions of corresponding letters according to English alphabet.

**Example 1: B, C, A, D, Z, E, … F, X, G
**(a) U (b) Y

(c) W (d) V

Solution. (b) The sequence consists of two series B, A, Z, Y, X and C, D, E, F, G. The missing letter is Y.

**Example 2: P, U, Z, … J, 0, T
**(a) E (b) U

(c) S (d) P

Solution. (a) The sequence is P+ 5, U+ 5,Z+ 5. The missing letter is Z + 5 = E

**Example 3: B, D, G, I, … N, Q, S
**(a) I (b) J

(c) L (d) K

Solution. (c) The sequence is B + 2, D+ 3, G + 2, I + 3 and so on.

## Letter Series - Type 12

Two Letter Series The first letters of the series follow one logic and the second letters follow another logic.

**Example 4: EZ, DX, CV,..., AR, ZP
**(a) CS (b) AM

(c) BT (d) TG

Solution. (c) First and second letters follow a sequence of-1 and -2 respectively.

**Example 5: DG, HK, LO, PS, TW,…
**(a) XA (b) ZA

(c) XB (d) None of these

Solution. (a) First and second letters follow a sequence of + 4.

**Example 6: DX, EY FV, ... : ; HT, IU
**(a) HV (b) IX (c) GW (d) BZ

Solution. (c) First, -third and fifth terms follow a sequencee and second, fourth and sixth terms follow another sequence. (DX, FV, HT, etc) and (EY, GW, IU, etc).

## Letter Series - Type 3

Three Letter Series: :Such series consist of three letters in each term. The
first letters follow one logic, the second letters follow another logic and the
third letters follow some other logic.

Example 7: DIE, XCY, RWS, ...

(a) LQN (b) QMP

(c) LMS (d) LQM

Solution. (d) First, second and third letters of each group follow a sequence of
-6 series.

**Example 8: VPG, UQF, ..., SSD, RTC
**(a) SQD (b) TRE

(c) TRS (d) QDT

Solution. (b) First, second and third letters follow a sequence of –1, + 1, –1 series respectively.

**Example 9: DJS, HNW, LRA, PVE, ..., XDM
**(a) TZI (b) SAF

(c) UXH (d) None of these

Solution. (a) First, second and third letters follow a sequence of + 4 series.

## Letter Series - Type 4

A series of letters is given with one or more missing letters. From the choices, the choice that gives the letters that go into the blanks has to be selected as answer.

**Example 10: In the following series some letters are missing. From the
choices, select the choice that gives that letters that can fill the blanks in
the given sequence.
**a_ c_ b_ab_a_ca_c

(a) abaccb (b) accbab

(c) aabbcc (d) baccbb

Solution. (d) First of all, notice that there are 6 blanks in the given sequence and each choice gives six letters to fill the six blank in order. Now, we have to select an alternative which if placed in the blanks of the series in order, we get a complete series of letters which follow some particular pattern.

The best way is to try with each option. Inserting the letters of option (d) in place of the blanks, we get a series like “abc abc abc abc abc” which is a repetition of the group of letters “abc”.

## Letter Series - Type 5

Here, students are asked to count how many times a particular letter or group of letters satisfying some conditions occurs and mark that number as the answer choice. Example 11: In the following sequence of letters, in how many instances the letters n is immediately preceded by the letter t ?

s n r u a t n n g h j t k n s t n d g c l n t t t n n n t n t n t s m v b t n g c x d p t n k l s t n t

a) 5 b) 6 c) 7 d) 8

Solution. (d)On counting, we find that the letter n occurs 8 times, where n is immediately preceded by the letter t.

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