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Which of the following pairs replaces ? in series given. B4U, F12P, ?, N108FOPTIONS J19K J20Q J36K K36Q

Question

Which of the following pairs replaces ? in series given.

B4U, F12P, ? , N108F
OPTIONS
J19K
J20Q
J36K
K36Q

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Solution

To find the missing term in the series B4U, F12P, ?, N108F, we can break down the problem into smaller segments for analysis.

1. ### Break Down the Problem

We need to identify the pattern in the series and determine the missing pair by finding the connection in characters and numbers.

2. ### Relevant Concepts

The series consists of terms formed by two letters, a number, and then a final letter (i.e., Letter1NumberLetter2).

  1. The first letters in the series: B, F, ?, N
  2. The numbers: 4, 12, ?, 108
  3. The last letters: U, P, ?, F

3. ### Analysis and Detail

  • First Letters: The first letters are B, F, ?, N. Observing the positions in the alphabet:

    • B = 2
    • F = 6
    • N = 14
    • From B to F, there's a difference of 4 (6 - 2).
    • From F to N, there's a difference of 8 (14 - 6).
    • Thus, the third letter should be at position 2 + 4 + 8 = 14, which corresponds to the letter J.
  • Numbers: The sequence of numbers is 4, 12, ?, 108.

    • 4 is 4=4×14 = 4 \times 1
    • 12 is 12=4×312 = 4 \times 3
    • 108 can be viewed as 108=4×27108 = 4 \times 27.
    • The numbers appear to be successive multiples of 4, but it looks like they also grow by a multiplier that follows 1, 3, and then by 9 in the last known term. The missing term should be 4×7=284 \times 7 = 28.
  • Last Letters: Observing the last letters U, P, ?, F:

    • The positions in the alphabet are:
      • U (21), P (16), ?, F (6)
    • The difference between U and P is -5, and between P and F is -10, suggesting a consistent decrease.
    • Therefore, the middle term would be a letter that is 16 - 5 = 11, which corresponds to K.

4. ### Verify and Summarize

The identified missing term is thus J28K, considering the letter patterns and number sequences.

Final Answer

J28K

This problem has been solved

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