9.1: Fixed Skip Cyclic Patterns

 9.1: Fixed Skip Cyclic Patterns Alphabetical Reasoning Problems

1: Fixed Skip Cyclic Patterns

These problems involve a sequence where letters progress with a fixed skip (forward or backward) in the alphabet, wrapping around when the position exceeds 26 (Z) or goes below 1 (A).

Problems and Solutions

Problem 1: A, D, I, O, ?

• Step-by-Step Solution:
  1. List the positions: A = 1, D = 4, I = 9, O = 15.
  2. Calculate differences: 4 - 1 = 3, 9 - 4 = 5, 15 - 9 = 6.
  3. The differences (3, 5, 6) are not constant, but check the sequence: 3 + 2 = 5, 5 + 1 = 6. The next difference may be 6 + 0 = 6 (constant skip).
  4. Add 6 to O’s position (15): 15 + 6 = 21 (U).
  5. The next term is U.
• Answer: U
• Explanation: The sequence has a stabilizing skip of 6 positions (5 letters). From O, skipping 5 letters (P, Q, R, S, T) gives U. No modular wrapping is needed here, but the pattern is cyclic in its consistent skip.

Problem 2: B, I, S, I, ?

• Step-by-Step Solution:
  1. Positions: B = 2, I = 9, S = 19, I = 9.
  2. Differences: 9 - 2 = 7, 19 - 9 = 10, 9 - 19 = -10.
  3. The pattern alternates: +7, +10, -10. Assume the next difference is +7 (cyclic repetition).
  4. Add 7 to I’s position (9): 9 + 7 = 16 (P).
  5. The next term is P.
• Answer: P
• Explanation: The sequence follows a cyclic pattern of differences (+7, +10, -10). After I, the next skip is +7, giving P.

Problem 3: Z, W, I, U, ?

• Step-by-Step Solution:
  1. Positions: Z = 26, W = 23, I = 9, U = 21.
  2. Differences: 23 - 26 = -3, 9 - 23 = -14, 21 - 9 = 12.
  3. Check for a fixed backward skip: From Z to W, skip 2 letters backward (Y, X). From W to I, skip 13 letters backward.
  4. Assume a fixed backward skip of 3 positions: From U (21), subtract 3: 21 - 3 = 18 (R).
  5. The next term is R.
• Answer: R
• Explanation: The sequence may stabilize with a fixed backward skip of 3 positions (2 letters). From U, skip Q, R to reach R.

Problem 4: C, K, I, Z, ?

• Step-by-Step Solution:
  1. Positions: C = 3, K = 11, I = 9, Z = 26.
  2. Differences: 11 - 3 = 8, 9 - 11 = -2, 26 - 9 = 17.
  3. Check for a fixed forward skip: From I to Z, the skip is 17. Assume the pattern stabilizes.
  4. Add 17 to Z’s position (26): 26 + 17 = 43. Subtract 26: 43 - 26 = 17 (Q).
  5. The next term is Q.
• Answer: Q
• Explanation: The sequence may follow a large forward skip (17 positions). From Z, skip 16 letters, wrapping around to Q.

Problem 5: A, G, O, I, ?

• Step-by-Step Solution:
  1. Positions: A = 1, G = 7, O = 15, I = 9.
  2. Differences: 7 - 1 = 6, 15 - 7 = 8, 9 - 15 = -6.
  3. The pattern alternates: +6, +8, -6. Assume the next is +8 (cyclic).
  4. Add 8 to I’s position (9): 9 + 8 = 17 (Q).
  5. The next term is Q.
• Answer: Q
• Explanation: The sequence cycles through differences (+6, +8, -6). After I, the next skip is +8, giving Q.

Problem 6: Y, I, U, I, ?

• Step-by-Step Solution:
  1. Positions: Y = 25, I = 9, U = 21, I = 9.
  2. Differences: 9 - 25 = -16, 21 - 9 = 12, 9 - 21 = -12.
  3. The pattern alternates: -16, +12, -12. Assume the next is +12.
  4. Add 12 to I’s position (9): 9 + 12 = 21 (U).
  5. The next term is U.
• Answer: U
• Explanation: The sequence cycles through differences (-16, +12, -12). After I, the next skip is +12, giving U.

Problem 7: F, Q, I, I, ?

• Step-by-Step Solution:
  1. Positions: F = 6, Q = 17, I = 9, I = 9.
  2. Differences: 17 - 6 = 11, 9 - 17 = -8, 9 - 9 = 0.
  3. The repetition at I suggests a pause. Assume a fixed skip from I (e.g., +11 as from F to Q).
  4. Add 11 to I’s position (9): 9 + 11 = 20 (T).
  5. The next term is T.
• Answer: T
• Explanation: The sequence may cycle with a forward skip of 11 positions after the pause at I, giving T.

Problem 8: X, I, I, Z, ?

• Step-by-Step Solution:
  1. Positions: X = 24, I = 9, I = 9, Z = 26.
  2. Differences: 9 - 24 = -15, 9 - 9 = 0, 26 - 9 = 17.
  3. Assume a fixed forward skip after the pause: From I to Z, the skip is 17.
  4. Add 17 to Z’s position (26): 26 + 17 = 43. Subtract 26: 43 - 26 = 17 (Q).
  5. The next term is Q.
• Answer: Q
• Explanation: The sequence cycles with a forward skip of 17 positions after the pause at I, giving Q.

Problem 9: B, I, Q, I, ?

• Step-by-Step Solution:
  1. Positions: B = 2, I = 9, Q = 17, I = 9.
  2. Differences: 9 - 2 = 7, 17 - 9 = 8, 9 - Fits into a pattern where the skip stabilizes.
  3. Assume a fixed skip of 7 (as from B to I): From I (9), add 7: 9 + 7 = 16 (P).
  4. The next term is P.
• Answer: P
• Explanation: The sequence cycles with a forward skip of 7 positions, giving P after I.

Problem 10: Z, I, U, I, ?

• Step-by-Step Solution:
  1. Positions: Z = 26, I = 9, U = 21, I = 9.
  2. Differences: 9 - 26 = -17, 21 - 9 = 12, 9 - 21 = -12.
  3. The pattern alternates: -17, +12, -12. Assume the next is +12.
  4. Add 12 to I’s position (9): 9 + 12 = 21 (U).
  5. The next term is U.
• Answer: U
• Explanation: The sequence cycles through differences (-17, +12, -12). After I, the next skip is +12, giving U.