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Difficulty: HardNumber and Alphabet Series
Consider the following alphanumeric sequence:
C3Z,E7Y,H16V,M35Q,T74J,…C3Z, E7Y, H16V, M35Q, T74J, \dots
What is the next term in this sequence?
  1. E153AE153AAnswer
  2. B
    E153BE153B
  3. C
    F153AF153A
  4. D
    E154AE154A

Answer

The next term in the sequence is E153AE153A.
The term E153AE153A is derived by combining three distinct concurrent rules: adding consecutive prime numbers (+2, +3, +5, +7, +11) to the first letter's position, applying the operations ×2+1,×2+2,…\times 2 + 1, \times 2 + 2, \dots to the numeric value, and subtracting consecutive odd numbers (-1, -3, -5, -7, -9) from the last letter's position.

Step-by-Step Solution

1
Analyze the pattern of the first letter in each term
The positional values of the first letters are C=3,E=5,H=8,M=13,T=20C=3, E=5, H=8, M=13, T=20. The differences between consecutive terms are +2,+3,+5,+7+2, +3, +5, +7. These differences are prime numbers. The next prime number to add is +11+11. T(20)+11=31T(20) + 11 = 31. Subtracting 26 for alphabetical wraparound gives 31−26=531 - 26 = 5, which corresponds to the letter EE.
To establish the rule governing the first component of each term.
2
Analyze the numerical pattern in the middle of each term
The numerical sequence is 3,7,16,35,743, 7, 16, 35, 74. The recurrence relation is Termk+1=Termk×2+k\text{Term}_{k+1} = \text{Term}_k \times 2 + k. Specifically: 3×2+1=73 \times 2 + 1 = 7, 7×2+2=167 \times 2 + 2 = 16, 16×2+3=3516 \times 2 + 3 = 35, 35×2+4=7435 \times 2 + 4 = 74. The next term is 74×2+5=148+5=15374 \times 2 + 5 = 148 + 5 = 153.
To identify the algebraic multiplier and incremental addition rule of the numeric component.
3
Analyze the pattern of the second letter in each term
The positional values of the second letters are Z=26,Y=25,V=22,Q=17,J=10Z=26, Y=25, V=22, Q=17, J=10. The differences between consecutive terms are −1,−3,−5,−7-1, -3, -5, -7. These subtractions follow consecutive odd numbers. The next odd number to subtract is −9-9. J(10)−9=1J(10) - 9 = 1, which corresponds to the letter AA.
To determine the positional shift pattern for the final letter element.
4
Combine the results from all three components
Combining the first letter (EE), the middle number (153153), and the second letter (AA) yields E153AE153A.
To form the complete missing term of the alphanumeric sequence.

Key Concept

Alphanumeric Series Completion with Interdependent Prime, Odd, and Linear Operations
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Topics

Number and Alphabet Series

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