In the alphabetic series CGE, RNP, FJH, OKM, IMK, __, find the next three letter group that correctly follows the pattern.

Introduction / Context:
This question features an alphabetic series of three letter groups where the pattern is not immediately obvious from one group to the next. Often such sequences are made by interleaving two simpler subsequences, each with its own rule. Our goal is to uncover these subsequences and then extend them to determine the missing group.

Given Data / Assumptions:

• Series: CGE, RNP, FJH, OKM, IMK, __.
• Each term is a group of three letters.
• The pattern may be split into odd and even positioned groups.

Concept / Approach:
The strategy is to separate groups at odd and even positions. The odd positioned groups (first, third and fifth) may follow one rule, while the even positioned groups (second and fourth, and then sixth) may follow another. After recognising these patterns, we use them to compute the next even positioned group.

Step-by-Step Solution:
Step 1: List odd position groups: 1st CGE, 3rd FJH, 5th IMK. Convert letters to positions: C=3, G=7, E=5; F=6, J=10, H=8; I=9, M=13, K=11.Step 2: Observe that from CGE to FJH each letter increases by 3 (3 to 6, 7 to 10, 5 to 8). From FJH to IMK, again each letter increases by 3 (6 to 9, 10 to 13, 8 to 11).Step 3: Thus, the odd subsequence advances by adding 3 to each letter. If we extended it further, the next odd group would be LPN, but that is not required for this question.Step 4: Now list even position groups: 2nd RNP and 4th OKM. Positions are R=18, N=14, P=16 and O=15, K=11, M=13.Step 5: From RNP to OKM each letter decreases by 3 (18 to 15, 14 to 11, 16 to 13).Step 6: Continue this rule: subtract 3 from IMK (the 5th group) to find the 6th group. I=9 becomes F=6, M=13 becomes J=10, K=11 becomes H=8, so we get FJH. However FJH is not an option, so we recognise that the even subsequence should be derived from the previous even group OKM, not from IMK.Step 7: Apply the subtraction rule to OKM instead: O=15 minus 3 = L=12, K=11 minus 3 = H=8, M=13 minus 3 = J=10, so the next even group is LHJ.

Verification / Alternative check:
Now the two subsequences are: odd groups CGE, FJH, IMK (each +3 per letter) and even groups RNP, OKM, LHJ (each -3 per letter). The interleaved full series becomes CGE, RNP, FJH, OKM, IMK, LHJ, which is coherent. Among the options, only LHJ fits this dual pattern.

Why Other Options Are Wrong:
Groups such as JHL, HJL or JKL do not arise from subtracting 3 from all letters of OKM. At least one letter in each of these alternatives breaks the consistent minus 3 progression of the even subsequence, so they cannot be the correct continuation.

Common Pitfalls:
One pitfall is to try to find a single rule relating each group to the next in sequence, which is very complicated here. Recognising that two subsequences are interleaved simplifies the task drastically. Another error is to change the direction of movement for some letters but not others, which destroys the regularity of the pattern.

Final Answer:
The three letter group that correctly continues the series is LHJ.