# Letter and Symbol Series Questions and Online Test

In this type of test, a series of numbers or alphabetical letters are given in questions. The given series follows a certain rule and pattern throughout in given set. Therefore, you are required to recognize the pattern or rule and to complete the given series or find out the wrong one in the series.

## Number Series

In this type of test, a number series is given, out of which one number or alphabet is missing. So you are required to complete the series by analyzing the applied rule or pattern.

## Rule for the Number Series

Some of the standard rules followed in estimating the next term are given below:

1. Each number may be the multiple of the other.
2. In each number something may be added or deducted to get the second number and so on.
3. The numbers may be divided to get the next number.
4. The number may be squared to get the next one.
5. The square root of the numbers may be taken to get the next number.
6. Sometimes, odd numbers are followed or preceded by even numbers and vice versa.
7. Sometimes, sometimes is added to get the next number and subtracted or divided or multiplied to get the next one.
8. Sometimes, odd number is squared, the other multiplied thrice, fourth multiplied four times, etc. and then sometimes is added or subtracted.

## Solved Example - Number Series

Question 1. 20, 32, 45, 59, 74, ?
(A) 95
(B) 90
(C) 85
(D) 79

Solution: The differences between the consecutive terms of the series are 12, 13, 14, 15, 16, …..
2nd term = 1st term + 12 = 20 + 12 = 32
3rd term = 2nd term + 13 = 32 + 13 = 45
4th term = 3rd term + 14 = 45 + 14 = 59
5th term = 4th term + 15 = 59 + 15 = 74
Thus, ? = 6th term = 5th term + 16 = 74 + 16 = 90
Hence, the answer is (B).

Question 2. 210, 195, 175, 150, 120, ?
(A) 75
(B) 80
(C) 85
(D) 90

Solution: The differences between the consecutive terms of the series are -15, -20, -25, -30, -35, …
2nd term = 1st term – 15 = 210 – 15 = 195
3rd term = 2nd term – 20 = 195 – 20 = 175
4th term = 3rd term – 25 = 175 – 25 = 150
5th term = 4th term – 30 = 150 – 30 = 120
Thus, ? = 6th term – 35 = 120 – 35 = 85
Hence, the answer is (C).

Question 3. 3, 5, 10, 12, 24, 26, ?
(A) 52
(B) 30
(C) 28
(D) 48

Solution: Each odd term doubles the previous even term.
3th term = 2×2nd term=2×5=10
5th term = 2×4th term=2×12=24
Thus, = 7th term = 2×6th term=2×26=52
Hence, the answer is (A).
Hence, the answer is (A).

Question 5. 2, 6, 10; 5, 9, 13; 10, 14, 18; ?, ?, ?.
(A) 12, 14, 16
(B) 14, 17, 20
(C) 17, 21, 25
(D) 15, 20, 25

Solution: Each group have a common difference of 4 among its consecutive terms. Also the difference between the consecutive first term of each group is 3, 5, 7, etc.
Thus, ?, ?, ? = ( 10 + 7), (17 + 4), (21 + 4)
= 17, 21, 25,
Hence, the answer is (C).

## Wrong Number

In this type, some numbers are arranged in a series and in sequence. One term in the series is wrong. You are asked to find out the wrong term, from the alternatives given. Therefore, you have to decipher a general pattern followed in the sequence, and then decide the wrong term accordingly.

## Solved Example - Wrong Number

Question 1. 1, 11, 26, 49, 80, 122, 169, 227, 290, 361
(A) 49
(B) 26
(C) 80
(D) 122

Solution: The differences between the consecutive terms of the series are
5, 8, 11; 5, 8, 11; ………
∴ 2nd term = 1st term + 10 = 1 + 10 = 11
3rd term = 2nd term + 10 + 5 = 11 + 10 + 5 = 26
4th term = 3rd term + 10 + 5 + 8 = 26 + 10 + 5 + 8 = 49
5th term = 4th term + 10 + 5 + 8 + 11 = 49 + 10 + 5 + 8 + 11 = 83
Thus, fifth term should be 83 and not 80.
Hence, the answer is (C).

Question 2. 11, 5, 20, 12, 40, 26, 74, 54
(A) 5
(B) 20
(C) 26
(D) 40

Solution: The differences between odd terms are 9, 18, 36, ……….
3rd term = 1st term + 9 = 11 + 9 = 20
5th term = 3rd term + 18 = 20 + 18 = 38
7th term = 5th term + 36 = 38 + 36 = 74
Thus, 5th term should be 38 and not 40.
Hence, the answer is (D).

Question 3. 3, 6, 5, 20, 7, 42, 9, 74
(A) 6
(B) 42
(C) 74
(D) 20

Solution: Differences between the even terms of the series are 14, 22, 30……
4th term = 2nd term + 14 = 6 + 14 = 20
6th term = 4th term + 22 = 20 + 22 = 42
8th term = 6th term + 30 = 42 + 30 = 72
Thus, 8th term should be 72 and not 74.
Hence, the answer is (C).

Question 4. 19, 17, 14, 11, 7
(A) 14
(B) 17
(C) 11
(D) 19

Solution: The difference between odd terms of the series is 6.
3rd term = 1st term – 6 = 19 – 6 = 13
5th term = 3rd term – 6 = 13 – 6 = 7
Thus, 3rd term should be 13 and not 14.
Hence, the answer is (A).

## Alphabet Series

In such series, a series of alphabets is given, and which follow a pattern throughout in a series. Therefore, you have to decipher the pattern and complete the series, by choosing the correct alternative given in question.

## Rules for the Alphabet Series

Some of the standard rules followed in estimating the next term in alphabet series are discussed below:

1. Keep the order of letters with their respective numbers, i.e. A = 1 …Z = 26 and vice versa i.e. Z = 26 and vice versa i.e. Z = 1 … A = 26.
2. When the counting is circular, after Z, the cycle will continue from A.
3. Sometimes, letters are committed in sequence, e.g. ABC-, EFG-, Here D and H in sequence are omitted.
4. Sometimes, equal number of letters are omitted, e.g. in ADG, two consecutive letters are omitted.
5. One following the letter and the other preceding it may be organized e.g. in BCA, C follows B and A precedes B.
6. Sometimes, there may be repetition of letters in a set order, e.g. in aab, abc-, one letter is repeated twice, the next set could be ccd.
7. Sometimes, numbers are mixed with letters, the number may refer to the position of the letters in the alphabet.

## Solved Example - Alphabet Series

Question 1. ABXW, EFTS, ?, MNLK
(A) IJOP
(B) IJPO
(C) JIOP
(D) JIPO

Solution: First and second terms are in alphabetical order, while third and fourth term are in reverse order. Therefore, missing item is IJPO.
Hence, the answer is (B).

Question 2. A Z X B V T C R ?
(A) E, O
(B) O, Q
(C) F, O
(D) P, D

Solution: The first, fourth and seventh letters are in alphabetical order. Thus, tenth letter would be after C i.e. D. Also, second and third letters are in reverse order and so are the fifth and sixth letters.
Hence, the answer is (D).

Question 3. P3C, R5F, T8I, VI2, L, ?
(A) YI7O
(B) XI7M
(C) XI7O
(D) XI6O

Solution: The first letter of the terms are alternating. And the sequences is followed by the numbers is +2, +3, +4 …….. The last latter of the each term has moved three steps forward to the last letter, ? = XI7O
Hence, the answer is (C)

Question 4. AK, EO, IS, ? , QA, UE
(A) LV
(B) MW
(C) NX
(D) LW
(E) MV

Solution: The first and second letters of each term are skipped four steps forward to obtain the next term. ? = MW.
Hence, the answer is (B).

## Symbol Series Questions from Previous Year Exams

Letter and Symbol Series

## Symbol Series Video Lecture

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