The document discusses different types of series questions that are designed to test logical patterns and relationships. There are three main types: 1. Number series - Questions involve identifying the pattern between numbers in a sequence to determine missing numbers. Patterns can be based on differences, products, powers, or other logical rules. 2. Letter series - Questions replace letters with their position in the alphabet to identify patterns between letter sequences. Patterns may alternate between two series. 3. Alphanumeric series - Questions combine letters, numbers, and symbols to test pattern identification skills. The document provides examples and approaches for solving various series questions. Common approaches involve observing patterns, finding common differences, and identifying if a series follows multiplication,

1. Series

4. Series
• Series
1. Number & Letter series
• Concepts: Series are designed to measure a student’s ability to find logical patterns,
which may be based on simple mathematical and logical rules and properties. We have
3 types of series.
•
A. Number Series: A Sequence of numbers is given, where one needs to identify the
pattern between the numbers. This pattern needs to be applied in the series to find out
the missing number. There can be infinite ways to create a series, but all the numbers
within a series should have the same relationship between them.
•
• Important - Due to the various patterns that can be used to create a number series; this
can often be a difficult and time consuming question type. To solve problems on number
series, one should be familiar with the basics of numbers i.e. multiplication tables,
squares, cubes, powers, factorials etc.

5. • There are some important types of number series, which we
must be familiar with
• i. Simple observation- sometimes, the series is very simple &
can be solved by just observation. It is useful when the series is
an Arithmetic or Geometric Progression or series based on
prime numbers.
• For example-
• 1. 2, 5, 8, 11, …
• 2. 2, 6, 18, 54, …
• 3. 2, 3, 5, 7, …
• Now the series shown above are as
• 1. Arithmetic progression with 2 & 3 as 1st term & common
difference
• 2. It is an geometric series with 2 & 3 as 1st term & common
ratio
• 3. It is a series based on prime numbers

6. • ii. Based on Difference
• This is the most basic and common form of series & majority of questions are from this type. It is based on the
difference between consecutive terms of the series which can be either constant or based on a Mathematical
pattern of it’s own. If the numbers obtained after the first level of substraction do not show any pattern among
them, take the difference of these numbers which is also called the common difference. The numbers obtained
after second level of substraction may now show a pattern. Continue this process till a pattern is observed.
Occasionally, the difference between the terms may be based on special numbers such as prime numbers,
factorials, powers or roots.
• Important- Method of common difference is most important for solving number series. We can find common
difference for a square power series follow Arithmetic progression. Similarly, common difference of common
difference of a cubic power series follow Arithmetic progression
• Example 1: 336,305,268,227,184,?
• The difference between the 1st and 2nd term is 336-305=31
• The difference between the 2nd and 3rd term is 305-268=37
• The difference between the 3rd and 4th term is 268-227=41
• The difference between the 4th and 5th term is 227-184 = 43
• The difference between any 2 consecutive terms are all consecutive prime numbers.
• Therefore the 6th term must be 184-47=137
• Example 2 : 1,3,6,10,15,?
• The difference between consecutive terms is 2,3,4,5 and therefore, the 6th term must be 15+6=21
• The above series can also be seen as 1,(1+2),(1+2+3),(1+2+3+4) and so on which again tells us that there could be
multiple logic used to form the same series.
• Example 3: 1, 4, 9, 16, …
• Here clearly it is an square series. We can observe that common difference forms an Arithmetic series with 3, 5, 7.
So, for next term is must be 9 & so next term is 9+16 = 25
• It is helpful if the series is like 0, 3, 8, 15, …
• The series is as Tn = n2-1. We may not be able to find it by common observation. But, by observing arithmetic
progression in common difference we can solve such cases.

7. • iii. Based on Product : A product based series is one where the pattern is identified by a
process of multiplication or division. Common patterns used in these problems are powers,
factorials and multiples. A feature of such problems is that the value of consecutive terms
increases/decreases quite sharply. However, first level subtraction may often help in identifying the
underlying pattern.
• Example 1 : 6,15,35,77,?
• The 1st term is 2x3, the 2nd term is 3x5, the 3rd term is 5x7 and the 4th term is 7x11 ie the series is
the product of consecutive prime numbers.
• The next term will be 11x13 = 143
• Example 2 : 6,3,3,4.5,9,?
• The 2nd term can be written as 1st term x 0.5, the 3rd term can be written as 2nd term x 1, the 4th
term can be written as 3rd term x 1.5 and so on.
• Required value will be 9x2.5 = 22.5
• iv) Power series: In power based problems, the increase in the value of consecutive terms will be
higher as compared to pure product based problems. Mostly we will get square or cubic series or
based on them. Power series on higher powers is rarely used.
• Example 1 : 2,6,30,230,?
• The given series can be written as (1^1+1), (2^2+2), (3^3+3), (4^4+4)
• Required answer will be 5^5+5 = 3130

8. v) Factorial series : The values of the factorial of the first few natural
numbers is used as the basis for this series.
Example 1 : 1,2,4,9,28, ?
The given series can be written as 0!+0, 1!+1, 2!+2, 3!+3, 4!+4
The answer therefore will be 5!+5 = 120+5 = 125
vi) Alternating Series : An alternating series is a combination of two or
more series. Each series can have different patterns applied to it and
then combined to form a series. In a combination of 2 or more series,
alternate terms follow the same pattern.
Example 1 : 3,4,7,9,11,14,15,?
As can be observed, there are 2 different series, one with a constant
difference of 4 (terms being 3,7,11 and 15) and the other with a
constant difference of 5(4,9,14…). The answer therefore will be the
next term of the 2nd series mentioned ie 19.
Important- For alternating series we are mostly given large number of
terms so can we can identify the pattern for both series which are
present in the given alternating series

9. • (Vii) Multiplication series
• Example 1: 5, 16, 49, 148, …….
• Solution: In this case ,
• 16 = (5 × 3) + 1
• 49 = (16 × 3) + 1
• 148 = (49 × 3) + 1
• So, missing term = (148 × 3) + 1 = 445
• Example 2: 4, 7, 23, 89, …….
• Solution: If the pattern is not clear in starting terms, we can try to find pattern in this case by focussing on last 2 given terms and this
way go on to find the pattern
• 89 = (23 × 4) - 3
• 23 = (7 × 3) + 2
• 7 = (4 × 2) – 1
• So, the pattern is (× 2 – 1), (× 3 + 2), (× 4 – 3). So, for last 2 terms we must get (× 5 + 4)
• So, the missing term i.e. 5th term = (4th term × 5) + 4 = 89 × 5 + 4 = 449
• (Viii) Fibonacci series
• In this series 1st term = 2nd term = 1 and the 3rd term onwards a term is obtained by adding last 2 terms
• So, the series is = 1, 1, 2, 3, 5, 8, 13, 21, …..
• So, 1st term = 2nd term = 1.
• 3rd term = 1st term + 2nd term = 1+1 = 2
• 4th term = 2nd term + 3rd term = 1+2 = 3
• 5th term = 3rd term + 4th term = 2 + 3 = 5
• Similarly, 8th term = 6th term + 7th term = 8 + 13 = 21
• Sometimes we may be given series like = 2, 3, 5, 8, 13, 21, …..
• (in this case 1st term and 2nd term are not equal to 1). The missing term = 13 + 21 = 34

10. Approach
• Step 1: Observation
• We must be able to identify – Arithmetic or geometric progression series, Power series, Fibonacci series or
Alternating series by observation
• 1. Arithmetic or Geometric progression can be observed easily
• 2. Power series- If terms are close to 1, 4, 9, 16, 25 etc then the series is square series or derived from it
• If terms are close to 1, 8, 27, 64, 125 etc then the series is cubic series or derived from it
• 3. if no. of terms given are large, then there are strong chances of Alternating series or Fibonacci series
• Step 2: Common difference
• If we cannot identify the series by observation, then we should look at common difference and series may be
based on common difference or it may give idea of series
• Step 3- We can then check for product series
• Step 4- At must look the possibilities for Multiplication series
• (For that we should try to find relation by last 2 given terms and then should check for 2nd last and 3rd last and so
on cases)
• Example- 1, 5, 21, 85, ?
• Solution: Looking at 21 and 85, we can see 84 = (21 × 4) + 1
• So, may be the logic is New term =( last term × 4) + 1
• (i.e. the series is multiplication series). Now we can check the logic for 5 and 21 and then for 1 and 5 (it is
applicable)
• Step 5- At last we look for possibilities for Miscellaneous series (i.e. the series based on some other logic than
discussed above)
• Example – 1, 14, 33, 84, ?
• Solution- 1 =
• Similarly, many other type of miscellaneous series can be formed, which are most difficult to solve

11. Some important points regarding approach for number series
• 1. Multiplication series and common difference
• We can solve many problems based on multiplication series by common difference approach too.
• For example: 2, 7, 22, 67, ?
• In this case nth term = (n-1th term × 3) + 1
• So, 5th term = (67 × 3) + 1 = 202
• But the common differences are 5, 15, 45. So, common difference between 5th term and 4th term = 3 × 45
= 135
• Hence, 5th term = 67 + 135 = 202
• But some multiplication series cannot be solved by using Common difference
• For example: 4, 7, 23, 89, ?
• In this case 89 = (23 × 4) – 3; 23 = (7 × 3) + 2; 7 = (4 × 2) – 1
• So, the pattern is (× 2 – 1), (× 3 + 2), (× 4 – 3). So, for last 2 terms we must get (× 5 + 4)
• So, the missing term i.e. 5th term = (4th term × 5) + 4 = 89 × 5 + 4 = 449
• But, for series 4, 7, 23, 89, ? The common differences are = 3, 16, 66. We cannot find the pattern by using
common difference in this case.
• Reason- We can see if the treatment is same for all the terms (like × 3 + 1 in the 1st example), then the
series can be solved by using common difference too. But, if the treatment is not same, it is not possible
• Check yourself for 5, 11, 34, 137, ?
• This is multiplication series and missing term is 686. This problem cannot be solved by common difference.
Try to check it yourself

12. Some important points regarding approach for number series
• 2. Power series and common difference
• We can solve many problems based on power series by common difference approach too.
• For example:
• 1. 1, 4, 9, 16, ?
• 2. 1, 8, 27, 64, ?
• 3. -1, 6, 25, 62, ?
• 4. 3, 6, 11, 18, ?
• In this case 1st and 4th problems are based on square series (4th is derived series as nth term = n^2 + 2)
• In this case 2nd and 3rd problems are based on cube series (3rd is derived series as nth term = n^3 - 2)
• For square series (like 1st problem) or series derived from it (like 4th problem), the common difference
forms an Arithmetic Progression (A.P.) and for cubic series (like 2nd problem) or series derived from it (like 3rd
problem), the common difference of common difference forms an Arithmetic Progression
• But some problems based on power series cannot be solved by using Common difference
• For example: -2, 12, 22, 70, ?
• In this case: 1st term = -2 = (1^3 - 3); 2nd term = 12 = (2^3 + 4); 3rd term = 22= (3^3 – 5); 4th term = 70 = (4^3
+6)
• So, clearly 5th term must be 5^3 – 7 = 118
• But, for series -2, 12, 22, 70, ? The common differences are = 14, 10, 48. We cannot find the pattern by
using common difference in this case.
• Reason- We can see if the treatment is same for all the terms in a power series specially in derived series
(i.e. in problem 1st the nth term = n^2, in 2nd problem the nth term = n^3, in problem 3rd the nth term = n^3 -
2 and in problem 4th the nth term = n^2 + 2), then the series can be solved by using common difference too.
But, if the treatment is not same, it is not possible

13. B. Letter Series : A Letter series generally uses the position of a letter in the
alphabet or some other property of letters such as vowels/consonants etc.
In questions on letters, one should replace the letter by it’s corresponding position in
the alphabet thereby making the pattern simpler to understand. Like number series, a
letter series can also have alternating patterns.
Example 1 : Find the next 2 terms of the series L,M,O,N,R,O…..
The position of the alphabets is 12,13,15,14,18,15
We observe that there are 2 alternating series ie 12,15,18…… and 13,14,15……
Hence the next 2 terms in the series should have positions 21 and 16 respectively. The
next 2 terms should be U and P respectively.
C. Alphanumeric Series: These include a combination of alphabets, numbers
and symbols. Questions can be asked individually or as part of a group question.
Example 1 : Find the missing term. 1k1,1M3,1Q7,1S9
The position of each alphabet is K:11, M:13, Q:17, S:19
We can see that the numbers on the side of the letter when taken together, represent
the position of the letter in the alphabet. Also, the numbers are consecutive prime
numbers starting from 11.
Hence, the next term should correspond to the next prime number ie 23. The letter at
the 23rd position of the alphabet is W and therefore the next term should be 2W3.

14. Exercise
• Directions (1-14): What will be the next term in the following number series?
• 1. 2 , 3 , 5 ,7, ….
• a). 9 b).11 c).13 d).15
• 2. 5 , 15, 45 , …………..
• a).150 b).135 c).120 d).214
• 3. 1, 2, 5, 4, 25, 6, 125, ….
• a).8 b).625 c).120 d).7
• 4. 1, 1, 2, 3, 5, 8, 13, ….
• a). 18 b).20 c).21 d).16
• 5. 1, 4, 13, 40, …
• a).67 b). 49 c).121 d).214
• 6. 1, 2, 4, 7, 11, …..
• a).13 b).14 c).15 d).16
• 7. 1, 3, 9, 27, ….
• a).45 b).65 c).81 d).91
• 8. 1, 1, 2, 6, 24, ….
• a). 42 b).20 c).96 d).120
• 9. -1, 6, 7, 18, 23, ….
• a).35 b). 36 c).37 d).38
• 10. 3, 5, 31, 59, …..
• a).101 b).119 c).151 d).165
•

15. • 11. 3 , 8 , 25 ,74, ….
• a). 221 b).311 c).326 d).None of these
• 12. 7 , 15, 28 ,59, …………..
• a).114 b).135 c).120 d).214
• 13. 2, 7, 22, 67, ….
• a).373 b).624 c).417 d).None of these
• 14. 5, 15, 75, 525, ….
• a).4725 b).9245 c).4775 d).5725
•
• Directions (15-17): What will be the next term in the following alphabetic series?
• 15. A, E, I, O, ….
• a). S b).U c).W d).Y
• 16. A, D , G, J, ….
• a).M b).N c).O d).P
• 17. B, C , D, F, ….
• a).G b).H c).I d).J
• Directions (18-20): What will be the next term in the following alpha-numeric series?
• 18. 1, A, 4, E, 7, I, 10, O, ….
• a). 13 b).S c).W d).14
• 19. 1K1, 1L2, 1N4, 1Q7,
• a). 1R8 b).1S9 c).2T0 d).2U1
• 20. BC5, E5, GK18, M13,…
• a).QS36 b).UV43 c).TU42 d).Q17

16. • 21. 7 , 15 , 32,…….. ,138, 281
• a). 57 b).67 c).77 d).87
• 22. 4, 6, 9 ,14, …………..
• a).16 b).18 c).20 d).23
• 23. 260, 216, 128, 108, 62, 54, …….., 27
• a).39 b).49 c).29 d).19
• 24. 0, 3, 8, 15, …….., 35, 48
• a).22 b).11 c).24 d).26
• 25. 2, -1, 5, 2, ….., 7, 17, 14
• a). 9 b).10 c).6 d).8
• 26. 6, 15 , 33, 69, …., 285
• a).137 b).141 c).143 d).147
• 27. 3, 1.5, 1.5, 2.25, 4.5, ?
• 28. 3, 5, 16, 62, 312, ?