Question

Difficulty: MediumNumber and Alphabet Series
Consider the following alphanumeric series:
B4,E9,I19,N39,T79,B4, E9, I19, N39, T79, \dots
Which of the following terms logically completes the series?
  1. A159A159Answer
  2. B
    Z159Z159
  3. C
    A158A158
  4. D
    Z158Z158

Answer

The term that logically completes the series is A159A159.
The letter sequence advances by increasing positional shifts (+3, +4, +5, +6, +7), bringing T (20) to 27, which wraps around to A. The number sequence follows nk=2nk1+1n_{k} = 2n_{k-1} + 1, transforming 79 into 159. Combining these yields A159A159.

Step-by-Step Solution

1
Analyze the pattern governing the letter component of the series.
Positions in the English alphabet: B is 2, E is 5, I is 9, N is 14, T is 20.
Differences between consecutive letter positions:
52=+35 - 2 = +3
95=+49 - 5 = +4
149=+514 - 9 = +5
2014=+620 - 14 = +6
Following this progression, the next shift must be +7+7.
Adding 7 to the position of T (20) gives 20+7=2720 + 7 = 27. Wrapping around the 26-letter alphabet (2726=127 - 26 = 1), the 1st letter is A.
The letter pattern increases the positional shift increment by 1 at each step.
2
Analyze the pattern governing the numerical component of the series.
The numerical terms are 4, 9, 19, 39, 79.
Evaluating the relation between consecutive terms:
4×2+1=94 \times 2 + 1 = 9
9×2+1=199 \times 2 + 1 = 19
19×2+1=3919 \times 2 + 1 = 39
39×2+1=7939 \times 2 + 1 = 79
Applying the same rule to 79:
79×2+1=15979 \times 2 + 1 = 159.
Each number is generated by doubling the preceding number and adding 1.
3
Combine the letter and numerical results.
The next term in the series is A159A159.
Joining the identified letter A with the calculated number 159 yields the complete next term.

Key Concept

Alphanumeric Series Completion via Increasing Positional Shift and Recurrence Progression
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