Letter Series/Alphabet Series Questions

e) O, M

Best way to solve these type of questions is by converting the alphabets series into number as follows

After mapping the alphabet series into numerical series, we can easily see that this is a decreasing series with a common difference of 2.

So the next two alphabets will be those who have numerical value as 15 and 13

• 25, 23, 21, 19, 17, 15, 13

Now, again convert number series to alphabet series

• 15 becomes ‘O’
• 13 becomes ‘M’

Hence Option e) O, M is correct.

• Find the missing letter in the given series:
• V, A, H, __
• a) M
• b) N
• c) O
• d) P
• e) Q

e) Q

Let’s first convert alphabet series into number series

V, A, H, ? becomes 22, 27, 34, ?

Now we can observe that the difference between

first and second number is 5,

second and third number is 7,

So the difference between third and fourth should be 9.

To get the missing number in the series,

add 9 to third number i.e, 34+9 = 43

so the series now becomes 22, 27, 34, 43

Now convert this number series into alphabet series

43 becomes Q

Our alphabet series becomes V, A, H, Q

Hence, Option e) Q is correct.

1. Select the letters that can replace the question marks (?) in the following series.
2. A, B, D, G, ?
3. a) M
4. b) L
5. c) K
6. d) H

1. c) K

1. Select the letters that can replace the question marks (?) in the following series.
2. A, C, F, H, ?, M
3. a) L
4. b) K
5. c) J
6. d) I

1. b) K

1. Find the missing letter in the following series
2. Z, S, W, O, T, K, Q, G, ?, ?
3. a) N, C
4. b) N, D
5. c) O, C
6. d) O, D

1. a) N, C

1. Select the letters that can replace the question marks (?) in the following series
2. Z, L, X, J, V, H, T, F, ?, ?
3. a) R, D
4. b) R, E
5. c) S, E
6. d) Q, D

1. a) R, D

1. Find the missing letters in the given series
2. b e d f ? h j ? l
3. a) i, m
4. b) m, i
5. c) i, n
6. d) j, m

1. a) i, m

• Select the letters that can replace the question marks (?) in the following series
• AZ, BY, CX, ?
• a) EF
• b) GH
• c) IJ
• d) DE
• e) DW

e) DW

First convert alphabet series into number series

AZ, BY, CX, ? becomes 1-26, 2-25, 3-24, ?-?

After converting to number series, split this series into two series as follows

Series I – 1, 2 , 3, ? 

Series II – 26, 25, 24, ?

Now these solve these two series separately

Series I – this is a progression series with common difference of 1, so the missing number becomes 3+1 = 4

1, 2, 3, 4

Series II- This is a decreasing series with common difference of 1, so the missing number becomes 24-1 = 23

26, 25, 24, 23

Now, merge these two series again, so the combined series becomes

1-26, 2-25, 3-24, 4–23

Now convert this number series into alphabet series

4-23 becomes DW

Our alphabet series becomes AZ, BY, CX, DW

Hence, Option e) DW is correct.

• Select the letters that can replace the question marks (?) in the following series
• BF, CH, ?, HO, LT
• a) DN
• b) EL
• c) EK
• d) EM
• e) FJ

c) Ek

Let’s first convert letter cluster series into number series

BF, CH, ?, HO, LT becomes 2-6, 3-8, ?-?, 8-15, 12-20

After converting to number series, split this series into two series as follows:

Series I – 2 , 3 , ? , 8 , 12

Series II – 6 , 8 , ? , 15 , 20

Now these solve these two series separately.

Series I: 2 , 3 , ? , 8 , 12

In this series, we can see the difference between the first and second number is 1 and the difference between last and second last is 4. So, we can notice that the difference between adjacent numbers is increasing as 1, 2, 3, 4.

Therefore, the difference between the second number i.e, 3 and the next number should be 2. So, the missing number becomes 3+2 = 5. Also, cross-check if the difference between the fourth and fifth term is 3.

8 – 5 = 3, Yes the difference is 3.

so the series I now becomes 2 , 3 , 5 , 8 , 12.

Series II: 6 , 8 , ? , 15 , 20

In this series also, we can see the difference between the first and second number is 2 and the difference between last and second last is 5. So, we can notice that the difference between adjacent numbers is increasing as 2, 3, 4, 5.

Therefore, the difference between the second number i.e, 8, and the next number should be 3. So, the missing number becomes 8+3 = 11. Also, cross-check if the difference between the fourth and fifth term is 4.

15 – 11 = 4, Yes the difference is 4.

so the series II now becomes 6, 8, 11, 15, 20

Now merge these two series again, so our combined series becomes

2-6, 3-8, 5–11, 8-15, 12-20

Now convert this number series into alphabet series

5-11 becomes EK

2-6, 3-8, 5–11, 8-15, 12-20 becomes BF, CH, EK, HO, LT

Our alphabet series becomes BF, CH, EK, HO, LT

Hence, Option c) EK is correct.

1. Select the letters that can replace the question marks (?) in the following series.
2. CE, GI, KM, OQ, ?
3. a) TW
4. b) TV
5. c) SU
6. d) RT
7. e) UW

1. c) SU

1. Select the letters that can replace the question marks (?) in the following series.
2. AD, EH, IL, ?, QT
3. a) LM
4. b) MN
5. c) MP
6. d) OM

1. c) MP

1. Find the missing letter in the following series
2. JE, LH, OL, SQ, ?
3. a) WV
4. b) WX
5. c) VW
6. d) VX
7. e) XW

1. e) XW

1. Select the letters that can replace the question marks (?) in the following series
2. CX, FU, IR, ?, OL, RI
3. a) LO
4. b) MN
5. c) NO
6. d) OP
7. e) OR

1. a) L, O

1. Find the missing letters in the given series
2. DF, GJ, KM, NQ, RT, ?
3. a) UW
4. b) YZ
5. c) XZ
6. d) UX
7. e) YA

1. d) UX

• Select the letters that can replace the question marks (?) in the following series
• OTE, PUF, QVG, RWH, ?
• a) SYJ
• b) TXI
• c) SXJ
• d) SXI
• e) TYJ

d) SXI

Convert this letter cluster series into number series as follows

OTE, PUF, QVG, RWH, ? becomes 15-20-5, 16-21-6, 17-22-7, 18-23-8, ?

After converting to number series, split this series into three series as follows

Series I – 15, 16, 17, 18, ?

Series II – 20, 21, 22, 23, ?

Series III – 5, 6, 7, 8, ?

Now these solve these three series separately

Series I – this is a progression series with a common difference of 1, so the missing number becomes 18+1 = 19

15, 16, 17, 18, 19

Series II- This is also an increasing series with a common difference of 1, so the missing number becomes 23+1 = 24

20, 21, 22, 23, 24

Series III- This is also an increasing series with a common difference of 1, so the missing number becomes 8+1 = 9

5, 6, 7, 8, 9

Now, merge these three series again, so the combined series becomes

15-20-5, 16-21-6, 17-22-7, 18-23-8, 19–24–9

Now convert this number cluster series into alphabet series

19–24–9 becomes SXI

Our alphabet series becomes OTE, PUF, QVG, RWH, SXI

Hence, Option d) SXI is correct.

1. Select the letter- cluster that can replace the question mark (?) in the following series.
2. CXB, HUI, MRP, ROW, ?
3. a) WLD
4. b) VKD
5. c) WLK
6. d) VKC

1. a) WLD

1. Select the letter- cluster that can replace the question mark (?) in the following series
2. EAC, GCE, IEG, ?
3. a) JHI
4. b) JGI
5. c) KGI
6. d) KHI
7. e) KIJ

1. c) KGI

1. Select the letter- cluster that can replace the question mark (?) in the following series
2. KQG, JTK, HXO, ECS, ?
3. a) BIX
4. b) AIW
5. c) BIV
6. d) AHW

1. b) AIW

1. Select the letter- cluster that can replace the question mark (?) in the following series
2. DAC, GWH, JSM, MOR, ?
3. a) PJV
4. b) QKV
5. c) PKW
6. d) QJW

1. c) PKW

1. Select the letter- cluster that can replace the question mark (?) in the following series
2. AOS, CQU, ?, GUY, IWA
3. a) FSX
4. b) GSW
5. c) ESW
6. d) DRV

1. c) ESW

• Select the letter-cluster that can replace the question marks (?) in the following series
• ACDE, FHIJ, KMNO, ?
• a) PRST
• b) PQRS
• c) QRST
• d) QSTU

a) PRST

Convert this letter cluster series into number series as follows

ACDE, FHIJ, KMNO, ?  becomes 1-3-4-5, 6-8-9-10, 11-13-14-15, ?-?-?-?

After converting to number series, split this series into four series as follows

Series I – 1, 6, 11, ?

Series II – 3, 8, 13, ?

Series III – 4, 9, 14, ?

Series IV – 5, 10, 15, ?

Now these solve these four series separately

Series I – this is a progression series with a common difference of 5, so the missing number becomes 11+5 = 16

1, 6, 11, 16

Series II- This is also an increasing series with a common difference of 5, so the missing number becomes 13+5 = 18

3, 8, 13, 18

Series III- This is also an increasing series with a common difference of 5, so the missing number becomes 14+5 = 19

4, 9, 14, 19

Series IV- This is also an increasing series with a common difference of 5, so the missing number becomes 15+5 = 20

5, 10, 15, 20

Now, merge these four series again, so the combined series becomes

1-3-4-5, 6-8-9-10, 11-13-14-15, 16–18–19–20

Now convert this number cluster series into alphabet series

16–18–19–20 becomes PRST

Our alphabet series becomes ACDE, FHIJ, KMNO, PRST

Hence, Option a) PRST is correct.

1. Select the letter-cluster that can replace the question marks (?) in the following series.
2. CKDL, EKFL, GKHL, _ , KKLL
3. a) IJKL
4. b) IKJL
5. c) MNOP
6. d) MNPQ

1. b) IKJL

1. Find the missing letter-cluster
2. AZCX, DWFU, GTIR, _ _ _ _
3. a) JRLO
4. b) JQLO
5. c) JRMO
6. d) JQMO

1. b) JQLO

1. Find the missing letter-cluster
2. ABMN, EFQR, IJUV, _ _ _ _
3. a) KLYZ
4. b) LMXY
5. c) MNYZ
6. d) LMYZ

1. c) MNYZ

1. Select the letter-cluster that can replace the question marks (?) in the following series
2. UVXY, PQST, KLNO, _ _ _ _, ABDE
3. a) FGHI
4. b) FGIJ
5. c) GHIJ
6. d) JIHG

1. b) FGIJ

1. Find the missing letter-cluster
2. AFKP, BGLQ, CHMR, _ _ _ _
3. a) DINS
4. b) EINS
5. c) DJOT
6. d) EJOT

1. a) DINS

• Select the letter-cluster that can replace the question mark (?) in the following series
• D-4, F-6, H-8, J-10, ?, ?
• a) K-12, M-13
• b) L-12, M-14
• c) L-12, N-14
• d) K-12, M-14

c) L-12, N-14

Our given series is D-4, F-6, H-8, J-10, ?, ?

Split the above series into two series i.e, the number and alphabet series as follows

Series I: Alphabet series – D, F , H, J, ?, ? 

Series II: Number series – 4, 6, 8, 10, ?, ?

Now these solve these two series separately

Series I – Convert this alphabet series into number series

D, F , H, J, ?, ? becomes 4, 6, 8, 10, ?,?

After converting we can easily identify the missing terms because this is a progression series with common difference of 2, so the missing terms are

10+2 = 12,

12+2=14

So the series becomes

4, 6, 8, 10, 12, 14

Now convert it again to alphabet series

4, 6, 8, 10, 12, 14 becomes D, F, H, J, L, N

Series II- this a progression series with common difference of 2, so the missing terms are

10+2 = 12,

12+2 = 14

So the series becomes

4, 6, 8, 10, 12, 14

Now, merge these two series again, so the combined series becomes

D-4, F-6, H-8, J-10, L-12, N-14

Hence, Option c) L-12, N-14 is correct.

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. 3F, 6G, 11I, 18L, ?
3. a) 21O
4. b) 25N
5. c) 27P
6. d) 27Q
7. e) 25P

1. c) 27P

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. KM5, IP8, GS11, EV14, ?
3. a) BX17
4. b) BY17
5. c) CY18
6. d) CZ17
7. e) CY17

1. e) CY17

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. J2Z, K4X, I7V, ? , H16R, M22P
3. a) I11T
4. b) L11S
5. c) L12T
6. d) L11T
7. e) L12S

1. d) L11T

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. 2Z5, 7Y7, 14X9, 23W11, 34V13, ?
3. a) 27U24
4. b) 47U15
5. c) 45U15
6. d) 47V14

1. b) 47U15

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. Q1F, S2E, U6D, W21C,?
3. a) Y66B
4. b) Y44B
5. c) Y88B
6. d) Z88B

1. c) Y88B

• Select the letter-cluster that can replace the question mark (?) in the following series
• AB, BA, ABC, CBA, ABCD
• a) ACBD
• b) BACD
• c) CABD
• d) DBAC
• e) DCBA

e) DCBA

In this question, first count the number of alphabets in one term

In above table, we can notice that the count of terms is increasing as 2, 2, 3, 3, 4, 4. So the missing term will have 4 alphabets.

Now try to observe the difference in adjacent terms,

Second term is written in reverse order of first.

We cannot found any relation between second and third term

Third and fourth terms have similar relationship as first and second terms

So we can divide the series in pairs as follows

AB, BA | ABC, CBA | ABCD, ?

Using the first and second pair relations, we can predict the relation between fifth and sixth term that is the Sixth term will reverse of fifth term that is DCBA.

Hence option e) DCBA is correct option.

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. AB, DEF, HIJK, ?, STUVWX
3. a) MNOPQ
4. b) LMNOP
5. c) LMNO
6. d) QRSTU

1. a) MNOPQ

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. AF, BGL, CHMR, ?
3. a) EJOTY
4. b) DINSX
5. c) DINSY
6. d) EJOTX

1. b) DINSX

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. X, VU, SRQ, ? , JIHGF
3. a) PONML
4. b) NMLK
5. c) ONML
6. d) NMKJ

1. c) ONML

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. A, FG, KLM, ? , UVWXY
3. a) PORS
4. b) PQRS
5. c) QRST
6. d) ORST

1. b) PQRS

1. Select the letter-cluster that can replace the question mark (?) in the following series
2. U, PQ, ? , FGHI, ABCDE
3. a) LMN
4. b) MNO
5. c) NOP
6. d) KLM
7. e) ONM

1. d) KLM

• Select the set of letters that when sequentially placed in the blanks of the given letter series will complete the series
• k _ lmml _ mk _ mmk _ lkkl _ m
• a) k,m, m, k, l
• b) l, k, m, k, k
• c) k, l, m, k, k
• d) k, l, k, l, m

d) k, l, k, l, m

Best way to solve to these question is fill the given blanks using each options one by one and see which option forms a particular pattern.

Option a) k,m, m, k, l

Let’s fill the blank spaces in the series with option (a)

k k lmml m mk m mmk k lkkl l m

We can notice that there is no particular logical pattern forming.

Option b) l, k, m, k, k

k l lmml k mk m mmk k lkkl k m

Using this option also no logical pattern is forming

Option c) k, l, m, k, k

k k lmml l mk m mmk k lkkl k m

Here, also no logical pattern forming

Option d) k, l, k, l, m

k k lmml l mk k mmk l lkkl m m

Here we can see a pattern of five letter is repeating in particular order.

kklmm | llmkk | mmkll | kklmm

Hence option d) k, l, k, l, m is correct.

• Select the set of letters that when sequentially placed in the blanks of the given letter series will complete the series
• _SWWS_WWWS_SWWWW_SSS
• a) W, S, S, S
• b) W, S, W, S
• c) W, W, S, S
• d) W, S, S, W

a) W, S, S, S

our given series is

_ SWWS _ WWWS _ SWWWW _ SSS

Now use every option one by one to identify which option forms logical pattern.

While looking for patterns try to form a group the letters. In general, patterns have of group of 2, 3, 4, and 5 in the series. In the previous example, it was a group of 5

Option a) Using this option our series becomes

W SWWS S WWWS S SWWWW S SSS

Now try to form group of 2, 3, 4, or 5 to identify pattern

We can notice a pattern in which the alternate letters are appearing with increasing group size as 2, 2, 3, 3, 4, 4.

W |S|WW|S S| WWW|S S S|WWWW| S SSS

Here we can see that a logical pattern is formed.

Hence option a) W, S, S, S is correct.

1. Select the set of letters that when sequentially placed in the blanks of the given letter series will complete the series
2. pq _ _ qpp _ _ pq _
3. a) ppppp
4. b) ppqpp
5. c) ppqpq
6. d) qppqq

1. b) ppqpp

1. Select the set of letters that when sequentially placed in the blanks of the given letter series will complete the series
2. aab _ ab _ cabcca _ bcab _ c
3. a) bbbc
4. b) bbab
5. c) cabc
6. d) cbab

1. d) cbab

1. Select the set of letters that when sequentially placed in the blanks of the given letter series will complete the series
2. c _ bba _ cab _ ac _ ab _ ca
3. a) abcbc
4. b) acbcb
5. c) babcc
6. d) bcacb

1. b) acbcb

1. Select the set of letters that when sequentially placed in the blanks of the given letter series will complete the series
2. b _ b _ bb _ _ bbb _ bb _b
3. a) bbbbba
4. b) bbaaab
5. c) ababab
6. d) aabaab

1. c) ababab

1. Select the set of letters that when sequentially placed in the blanks of the given letter series will complete the series
2. cab _ a _ c _ bc _ bc _ b _ ab
3. a) bcbbab
4. b) bcbbbc
5. c) acacab
6. d) cbaaac

1. d) cbaaac