5.1 - Identifying or Extending a Repeating Pattern

5.1 - Identifying or Extending a Repeating Pattern

This type involves recognizing a repeating pattern in a sequence and extending it to find specific terms or continue the sequence.

Problem 1

Question: The sequence A, B, A, B, A, B follows a pattern. What is the 10th letter?

Solution Process:

• Identify the pattern: A, B, A, B, A, B alternates between A and B.
• The repeating unit is AB (period = 2).
• For the nth letter: If n is odd, the letter is A; if n is even, the letter is B.
• n = 10 (even): 10th letter = B.

Solution: B

Detailed Explanation:

• The sequence alternates A, B, A, B, ..., with a period of 2 (AB repeats).
• Positions: 1=A, 2=B, 3=A, 4=B, ..., 9=A, 10=B.
• Since 10 is even, the 10th letter follows the pattern’s second term (B).
• This tests basic pattern recognition and indexing.

Problem 2

Question: The sequence A, B, C, A, B, C follows a pattern. What is the 8th letter?

Solution Process:

• Pattern: A, B, C, A, B, C has a repeating unit of ABC (period = 3).
• For the nth letter: Position n mod 3 determines the letter (1=A, 2=B, 3=C).
• n = 8: 8 ÷ 3 = 2 remainder 2 (8 mod 3 = 2). 2nd letter in cycle = B.

Solution: B

Detailed Explanation:

• The sequence repeats ABC every 3 letters: 1=A, 2=B, 3=C, 4=A, 5=B, 6=C, 7=A, 8=B.
• For n=8, compute 8 mod 3 = 2, so the 8th letter is the 2nd letter in the cycle (B).
• This tests modular arithmetic to find a term in a repeating sequence.

Problem 3

Question: The sequence X, Y, X, Y, X, Y follows a pattern. What are the next two letters?

Solution Process:

• Pattern: X, Y, X, Y, X, Y alternates XY (period = 2).
• Last given letter: 6th = Y.
• Next letters: 7th = X (odd position), 8th = Y (even position).

Solution: X, Y

Detailed Explanation:

• The sequence XY repeats: 1=X, 2=Y, 3=X, 4=Y, 5=X, 6=Y.
• After the 6th letter (Y), the pattern continues with the 7th (X) and 8th (Y).
• This tests the ability to extend a simple alternating pattern.

Problem 4

Question: The sequence P, Q, R, S, P, Q, R, S follows a pattern. What is the 12th letter?

Solution Process:

• Pattern: P, Q, R, S repeats every 4 letters (period = 4).
• For n=12: 12 ÷ 4 = 3 (no remainder). 12 mod 4 = 0, equivalent to 4th letter in cycle (S).
• Alternatively: 9=P, 10=Q, 11=R, 12=S.

Solution: S

Detailed Explanation:

• The sequence PQRS repeats: 1=P, 2=Q, 3=R, 4=S, 5=P, ..., 12=S.
• For n=12, since 12 mod 4 = 0, it corresponds to the 4th letter (S).
• This tests handling a longer repeating unit.

Problem 5

Question: The sequence A, A, B, A, A, B follows a pattern. What is the 9th letter?

Solution Process:

• Pattern: A, A, B repeats every 3 letters (period = 3).
• For n=9: 9 ÷ 3 = 3 (no remainder). 9 mod 3 = 0, equivalent to 3rd letter (B).
• Alternatively: 7=A, 8=A, 9=B.

Solution: B

Detailed Explanation:

• The sequence AAB repeats: 1=A, 2=A, 3=B, 4=A, 5=A, 6=B, 7=A, 8=A, 9=B.
• For n=9, 9 mod 3 = 0, so it’s the 3rd letter in the cycle (B).
• This tests recognizing a pattern with repeated letters.

Problem 6

Question: The sequence Z, Y, X, Z, Y, X follows a pattern. What is the 10th letter?

Solution Process:

• Pattern: Z, Y, X repeats every 3 letters (period = 3).
• For n=10: 10 ÷ 3 = 3 remainder 1. 10 mod 3 = 1 (1st letter = Z).
• Alternatively: 7=Z, 8=Y, 9=X, 10=Z.

Solution: Z

Detailed Explanation:

• The sequence ZYX repeats: 1=Z, 2=Y, 3=X, 4=Z, ..., 10=Z.
• For n=10, 10 mod 3 = 1, so it’s the 1st letter (Z).
• This tests a pattern with descending letters.

Problem 7

Question: The sequence A, B, A, C, A, D follows a pattern. What is the 7th letter?

Solution Process:

• Pattern: A, B (pair 1), A, C (pair 2), A, D (pair 3). The first letter is A, the second alternates B, C, D.
• Period = 2 (pairs). 7th letter is in the 4th pair (7 ÷ 2 = 3 remainder 1).
• Second letters: B (1st pair), C (2nd), D (3rd), E (4th). 7th letter (1st in 4th pair) = A.

Solution: A

Detailed Explanation:

• The sequence is paired: (A, B), (A, C), (A, D). The first letter is A; the second follows B, C, D, ...
• The 7th letter starts the 4th pair (A, E), so it’s A.
• This tests a pattern with a fixed letter and a progressing second letter.

Problem 8

Question: The sequence M, N, O, M, N, O follows a pattern. What are the next three letters?

Solution Process:

• Pattern: M, N, O repeats every 3 letters (period = 3).
• Last letter: 6th = O.
• Next letters: 7th = M, 8th = N, 9th = O.

Solution: M, N, O

Detailed Explanation:

• The sequence MNO repeats: 1=M, 2=N, 3=O, 4=M, 5=N, 6=O.
• After the 6th letter (O), the next cycle gives 7=M, 8=N, 9=O.
• This tests extending a three-letter repeating pattern.

Problem 9

Question: The sequence A, A, B, B, A, A, B, B follows a pattern. What is the 10th letter?

Solution Process:

• Pattern: A, A, B, B repeats every 4 letters (period = 4).
• For n=10: 10 ÷ 4 = 2 remainder 2. 10 mod 4 = 2 (2nd letter = A).
• Alternatively: 9=A, 10=A.

Solution: A

Detailed Explanation:

• The sequence AABB repeats: 1=A, 2=A, 3=B, 4=B, 5=A, ..., 10=A.
• For n=10, 10 mod 4 = 2, so it’s the 2nd letter (A).
• This tests a pattern with paired repetitions.

Problem 10

Question: The sequence X, Y, Z, X, Y, Z follows a pattern. What is the 15th letter?

Solution Process:

• Pattern: X, Y, Z repeats every 3 letters (period = 3).
• For n=15: 15 ÷ 3 = 5 (no remainder). 15 mod 3 = 0, equivalent to 3rd letter (Z).
• Alternatively: 13=X, 14=Y, 15=Z.

Solution: Z

Detailed Explanation:

• The sequence XYZ repeats: 1=X, 2=Y, 3=Z, 4=X, ..., 15=Z.
• For n=15, 15 mod 3 = 0, so it’s the 3rd letter (Z).
• This tests calculating a later term in a repeating pattern.