Difficulty: Medium
Correct Answer: NPT
Explanation:
Introduction / Context:
Triplet sequences frequently embed two fixed jumps. Recognizing the increment sizes (+2 then +3 here) helps isolate the single violator.
Given Data / Assumptions:
Concept / Approach:
Compute consecutive jumps per triplet and compare with the rule.
Step-by-Step Solution:
C(3)→E(5)=+2; E(5)→H(8)=+3 ✔K(11)→M(13)=+2; M(13)→P(16)=+3 ✔X(24)→Z(26)=+2; Z(26)→C(3)=(+3 with wrap) ✔N(14)→P(16)=+2; P(16)→T(20)=+4 ✖
Verification / Alternative check:
Recompute with a quick index list; only NPT breaks the +3 second jump.
Why Other Options Are Wrong:
CEH, KMP, XZC: All satisfy +2 then +3 (with cyclic wrap in XZC).
Common Pitfalls:
Forgetting cyclic wrap at Z, or miscounting steps by including the starting letter twice.
Final Answer:
NPT
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