Alphabet series with repeating step pattern – fill the last two letters: Z, Y, X, U, T, S, P, O, N, K, ?, ?

Introduction / Context:
Here the alphabet sequence mixes small and larger backward steps, repeating a short cycle. Recognizing the cycle and continuing it yields the final two letters.

Given Data / Assumptions:

• Sequence: Z, Y, X, U, T, S, P, O, N, K, ?, ?
• We must find the last two letters.

Concept / Approach:
Convert letters to positions and examine pairwise differences. A common motif is a cycle like −1, −1, −3 repeated throughout. Check whether this holds from the beginning and, if so, apply it to the tail.

Step-by-Step Solution:
Z(26) → Y(25): −1; Y(25) → X(24): −1; X(24) → U(21): −3.U(21) → T(20): −1; T(20) → S(19): −1; S(19) → P(16): −3.P(16) → O(15): −1; O(15) → N(14): −1; N(14) → K(11): −3.The cycle −1, −1, −3 repeats perfectly. Continue it: K(11) → J(10): −1; J(10) → I(9): −1.

Verification / Alternative check:
Mapping all steps confirms three complete repetitions of the −1, −1, −3 motif, and the final application gives J then I without any conflict.

Why Other Options Are Wrong:

• H, G or I, H break the exact step-by-step cycle visible across the entire sequence.
• H, I skips the necessary intermediate −3 that would follow after two −1 steps elsewhere in the pattern.

Common Pitfalls:

• Losing track of the repeating 3-step cycle and trying to invent a new rule at the end.

Final Answer:
J, I