J2Z, K4X, I7V, ?, H16R, M22P
2023
J2Z, K4X, I7V, ?, H16R, M22P
- A.
I11T
- B.
L11S
- C.
L12T
- D.
L11T
Attempted by 4 students.
Show answer & explanation
Correct answer: D
Treat each term as three independent sub-series: the FIRST letter, the MIDDLE number, and the LAST letter.
Series: J2Z, K4X, I7V, ?, H16R, M22P
1) Middle numbers: 2, 4, 7, ?, 16, 22. The differences increase by one each step: +2, +3, +4, +5, +6. So the missing number is 7 + 4 = 11 (and the next, 11 + 5 = 16, checks out).
2) Last letters: Z, X, V, ?, R, P, i.e. 26, 24, 22, ?, 18, 16. Each drops by 2, so the missing letter is 20 = T.
3) First letters alternate by position. Odd positions (1st, 3rd, 5th) are J, I, H = 10, 9, 8 (decreasing by 1). Even positions (2nd, 4th, 6th) are K, ?, M = 11, ?, 13 (increasing by 1), so the missing one is 12 = L.
Putting the three together, the missing 4th term is L11T.
Answer: L11T.