Triplet pattern decoding: J2Z, K4X, I7V, __, H16R, M22P — find the missing triplet by uncovering the independent rules for (1) the first letter, (2) the number, and (3) the last letter.

Introduction / Context:
Each term in the series is a triplet: (letter)(number)(letter). Such sequences commonly apply separate but synchronized rules to each component stream: the first-letter stream, the numeric stream, and the last-letter stream. The given series is: J2Z, K4X, I7V, __, H16R, M22P. We must determine the fourth triplet.

Given Data / Assumptions:

• First letters observed: J, K, I, ?, H, M.
• Numbers observed: 2, 4, 7, ?, 16, 22.
• Last letters observed: Z, X, V, ?, R, P.
• Alphabet positions: A=1 to Z=26.

Concept / Approach:
Last-letter stream: Z(26) → X(24) → V(22) → ? → R(18) → P(16): this is a steady −2 step each time, so the missing last letter is T(20).

Numeric stream: 2 → 4 (+2) → 7 (+3) → ? (+4) → 16 (+5) → 22 (+6). This is an increasing-difference pattern (+2, +3, +4, +5, +6). Therefore the missing number is 11 (7 + 4).

First-letter stream: J(10) → K(11) → I(9) → ? → H(8) → M(13). The deltas follow an alternating-signed, growing-magnitude pattern: +1, −2, +3, −4, +5. Applying +3 after I(9) yields L(12) for the fourth term; then −4 gives H(8), then +5 gives M(13), confirming the pattern.

Step-by-Step Solution:
1) Deduce last letter T via −2 steps from V (22) to T (20).2) Deduce number 11 via +4 from 7.3) Deduce first letter L via +3 from I.4) Assemble the triplet: L11T.

Verification / Alternative check:
Project forward and backward to ensure consistency: after L11T, the next given is H16R, which matches −4 on first letters (L→H), +5 on numbers (11→16), and −2 on last letters (T→R). Then H16R to M22P continues +5, +6, −2 as expected.

Why Other Options Are Wrong:
I11T has the wrong first letter (breaks the +3 rule); L11S and L12T break either the −2 last-letter rule (S ≠ T) or the numeric +4 step (12 ≠ 11).

Common Pitfalls:
Trying to force one global arithmetic rule across mixed components. Here, each of the three sub-sequences uses its own simple, consistent pattern.

Final Answer:
L11T