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Difficulty: MediumNumber and Alphabet Series

Which of the following terms logically completes the given alphanumeric series: C5X,F11U,I19R,L29O,…C5X, F11U, I19R, L29O, \dots?

  1. O41LO41LAnswer
  2. B
    O41MO41M
  3. C
    O39LO39L
  4. D
    N41LN41L

Answer

The term O41LO41L logically completes the given alphanumeric series.
The series follows three simultaneous patterns: the first letter advances by +3+3 positions (C→F→I→L→OC \to F \to I \to L \to O), the middle number increases by successive even integers +6,+8,+10,+12+6, +8, +10, +12 (5→11→19→29→415 \to 11 \to 19 \to 29 \to 41), and the last letter moves backward by −3-3 positions (X→U→R→O→LX \to U \to R \to O \to L). Combining these three components yields O41LO41L.

Step-by-Step Solution

1
Analyze the pattern of the first letter in each term.
The sequence of first letters is C(3),F(6),I(9),L(12)C (3), F (6), I (9), L (12).
The alphabetical positions increase by +3+3 each step. Thus, the next letter is at position 12+3=1512 + 3 = 15, which corresponds to OO.
2
Analyze the numerical progression in the middle of each term.
The numerical sequence is 5,11,19,295, 11, 19, 29.
The consecutive differences between terms are 11−5=611 - 5 = 6, 19−11=819 - 11 = 8, and 29−19=1029 - 19 = 10. The difference increases by +2+2 each step. The next difference is 10+2=1210 + 2 = 12, making the next number 29+12=4129 + 12 = 41.
3
Analyze the pattern of the third letter in each term.
The sequence of third letters is X(24),U(21),R(18),O(15)X (24), U (21), R (18), O (15).
The alphabetical positions decrease by −3-3 each step. Thus, the next letter is at position 15−3=1215 - 3 = 12, which corresponds to LL.
4
Combine the elements to find the missing term.
Combining the results yields O41LO41L.
Joining the first letter (OO), middle number (4141), and last letter (LL) completes the sequence.

Key Concept

Tri-pattern Alphanumeric Series Completion
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Topics

Number and Alphabet Series

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