Three-letter groups jumping by three – UGC NET Question

The starting letters of the groups are B, E, H and K, which occupy positions 2, 5, 8 and 11 in the alphabet, increasing by 3 each time. The groups themselves are made of three consecutive letters in forward order. Continuing the same rule, the next starting letter is position 14, which is N, and the consecutive letters after N are O and P. Therefore NOP is the only group that fits both the plus three jump in the starting letter and the three letter consecutive structure.

Option A:
Option A is correct because it begins with N, exactly three positions after K, and continues with O and P to form a consecutive three letter block. It preserves the constant increase of three positions in the first letter across all terms. It also maintains the same internal structure where each group is made of adjacent letters in correct order. These two properties together make NOP the valid continuation.

Option B:
Option B, MNO, starts at M, which is only two positions after K, so it breaks the fixed jump of three positions for the starting letters. Although MNO is itself a set of three consecutive letters, it does not arise from the same position rule used earlier. Because it violates the numerical pattern in the first letter, it cannot be the correct next term.

Option C:
Option C, OPQ, begins at O, which is four positions after K and overshoots the required step. This makes the gap between KLM and OPQ larger than all previous jumps. Since the series shows a stable plus three increment, the excessive jump in the starting letter makes OPQ inconsistent with the pattern.

Option D:
Option D, LMN, uses L as the starting letter, but L was already part of the previous group KLM and does not follow the plus three progression from K. It would make the sequence repeat an internal letter instead of moving to the next scheduled starting point. Therefore LMN fails to continue the series logically.