Letter, Number and continuous pattern Series with solved examples

Series

There are four types of series:

1.Letter series

2.Number series

3.Alphanumeric series

4.Continuous Pattern series

1.Letter Series:

In this series, only letters are available which follow certain patterns you have to find out that pattern and answer the questions.

A	1		26	Z
B	2		25	Y
C	3		24	X
D	4		23	W
E	5		22	V
F	6		21	U
G	7		20	T
G	8		19	S
I	9		18	R
J	10		17	Q
K	11		16	P
L	12		15	O
M	13		14	N

Positional values:

you have to learn the position of each letter. Along with the position of the opposite letter. To solve problems quickly.

To learn these values you can learn every letter with a positional difference of 5.

Then the word will be  EJOTY.

or you can make pairs of 5 letters 

ABCDE/FGHIJ/KLMNO/PQRST/UVWXY/Z

or you can make pairs of 3 letters

ABC/DEF/GHI/JKL/MNO/PQR/STU/VWX/YZ

or you can remember them by mapping them like a mobile keypad

or you can write them before the exam starts on the rough sheet like me.

A has positional value 1. and Z has positional value 26. But these values can be interchanged 

A becomes 26 and z becomes 1. 

To remember this you can make pairs like.

AZ, BY, CX DW, EV, FU, GT, IR, JQ, KP, LO, MN

Ex.1 BMO, EOQ, HQS,?

Sol. The first letter follows pattern B+3 = E +3 = H +3 = K

The second letter follows pattern   M+2 = O+3 = Q+2 = S 

The third letter follows patterns     O+2 = Q +2 = S+2 = U

KSU Ans.

Ex.2 C,Z,F,X,I,V,L,T,O,?,?

Sol.2 This series consists of two different series.

C+2 = F +2 = I +2 = L +2 = O +2 = R

Z -1 = X-1 = V -1 = T -1 = R

Thus R, R is the solution.

 

Ex.3ATTRIBUTION,TTRIBUTIO,RIBUTION,IBUTI,?

Sol.3 In first step one letter from beginning and one from the end of a term is removed.

 

In second step two letter from the beginning is removed. these two steps are repeated alternatively.

 

UTI Sol.

2.Number Series

Number series contains numbers only and you have to find the pattern and answer accordingly.

 

Patterns may include:

 

i. Addition series - Next term will be given  by addition of a particular number.

ii.Subtraction series - Next term will be given by subtraction of a particular number.

iii.Multiplication series - Next term will be given by multiplication of a particular number.

iv.Division series - Next term will be given by Division of a particular number.

 

v.Square series - Next term will be given by Square of a particular number.

 

vi.Cube series - Next term will be given by Square of a particular number.

 

vii.Triangular pattern series -  The differences between the consecutive terms of a series again form a series. the differences between the consecutive terms of the new series so formed, again from a series. This pattern continues till we attain a uniform difference between the consecutive terms of the series.

 

viii.Miscellaneous Series - next term will be given by combination of above series.

     for example , square and subtraction series, cube and addition series , division and multiplication     series.

All these will be cleared in solved examples.

Elementary idea of progression

i. Arithmetic progression : The progression of the form a, a+d, a+2d, a+3d is known as arithmetic progression

 

n^th term = a + (n-1)d

 

a= first term

d= common difference

n= nth term we want to find 

ii. Geometric progression : The progression of the form a, ar, ar², ar³ is known as GP.

a = first term

r = common ratio

n^th term = ar^n-1

Previous year Questions based on Number series :

Ex.1 14,17,21,24,28,31,?

Sol. This is an example of addition series.

14+3 = 17

17+4 = 21

21+3 = 24

24+4 = 28

28+3= 31

31+4= 35 Ans.

Ex.2 49,46,43,40, ?,34

Sol. This is an example of subtraction series.

49-3= 46

46-3= 43

43-3= 40

40-3= 37 Ans.

37-3= 34

Ex.3 9,18,72,576,?

Sol.This is multiplication series

9 × 2 =18

18 × 4 = 72

72 × 8 = 576

576 × 16 = 9216 Ans.

Ex.4 361,?,169,121,49,25

Sol.This is square series.

 

These numbers are in descending order of squares of prime numbers. 

19² = 361

17² = 289 Ans.

13² = 169

11² = 121

7² = 49

5² = 25 

Ex.5 2,5,12,27,?

Sol.this series is miscellaneous

2 ×(2+1)= 5

5 ×(2+2)=12

12×(2+3)=27

27×(2+4))= 58 Ans.

 

Ex.6 400,-200,100,?,25,-12.5 

Sol.This is division series

400 ÷ (-2) = -200

-200÷(-2)  = 100

100 ÷ (-2) = -50 Ans.

-50 ÷(-2) = 25

25 ÷(-2) = -12.5

Ex.7 2,6,4,9,8,13,16,18,32,?

Sol. This is  question is a combination of two series.

2,4,8,16,32 and 6,9,13,18,?

2 × 2 = 4

4 × 2 = 8

8 × 2 = 16

16×2 = 32

 

6 + 3 = 9

9 + 4 = 13

13+5 = 18

18 +6 = 24 Ans. 

Ex.8 1,8,27,64,125,216,?

Sol. This is the cube series.

1³ = 1

2³ = 8

4³ = 64

5³ =125

6³ = 216

7³ = 343 Ans.

Ex.9 5,10,13,26,29,58,61,122

Sol.

 

5×2 = 10

10+3 = 13

13×2 = 26

26+3 = 29

29 × 2 = 58

58+3 = 61

61×2 = 122 Ans.

Ex.10 0,6,24,60,120,210,? 

Sol. This is triangular series.

 

 

336  is the solution.

Ex.11 156,506,?,1806?

Sol.

1056 Ans.

Previous year Questions based on Alpha numeric series:

Ex.1 3F,6G,11I,18L,?

Sol. Numbers :  3+3 = 6 +5 = 11 +7 = 18 + 9 = 27  

 

Letters : F +1 = G +2 = I +2 = L +2 = P

 

27P Ans.

Ex.2 2Z5,7Y7,14X9,23W11,34V13,?

Sol 2. First Number series : 2 +5 = 7 +7 = 14 +7 = 23 +9 = 23 +11 = 34  +13 = 47

 

Middle letters : Z -1 = Y -1 = X -1 = W -1 = V -1 = U

 

Third number : 5 +2 = 7 +2 = 9 +2 =11 +2 =13 +2=15

 

47U15  Ans

4. Continuous pattern Series: It is a series of small letters which follows a certain pattern.However,

 Some letters are missing from the series. These missing letters are then given in a proper sequence as one of the alternatives.

Few examples based on continuous pattern series.

Ex.1 _ stt _ tt _ tts

 

(a). tsts    (b).ttst   (c). sstt    (d).tsst

 

Sol. (d) The series is tst/tst/tst/tst.. Thus,tst is repeated.

 

 

Ex.2 a _ n _ b_ _ ncb_ _ ncb 

 

(a)abbbcc   (b)abcbcb   (c). bacbab    (d).bcabab

Sol.2 (d)The series is abncb/abncb/abncb. Thus, the pattern abncb is repeated.