The groups descend in blocks of three letters: XYZ (24β26), UVW (21β23), RST (18β20) and OPQ (15β17). Each starting letter reduces by 3 positions (24, 21, 18, 15). The next starting position should be 12, corresponding to L, and the next block is LMN (12β14). Hence LMN is the only group that continues the pattern of moving each block three letters earlier in the alphabet.
Option A:
Option A, LMN, is correct because it begins at L, which is 3 letters before O, and forms the next three-letter block in descending order. It respects the length of the groups and the step size of β3 for the starting letter. This makes LMN a perfect fit for the series.
Option B:
Option B, KLM, starts too early at K, corresponding to a step of β4 from O, not β3. This over-adjustment does not match the observed behaviour of the series. As such, KLM cannot be considered correct.
Option C:
Option C, NOP, reverses the direction by moving forward after OPQ instead of backward. It also overlaps with OPQ by reusing O and P in a way that contradicts the blockwise descent. Therefore NOP does not align with the series rule.
Option D:
Option D, MNO, starts at M, which is 2 letters before O, not 3 letters before O. It breaks the fixed interval between the initial letters of successive blocks. Because it ignores the β3 step pattern, MNO is not a valid continuation.
Comment Your Answer
Please login to comment your answer.
Sign In
Sign Up
Answers commented by others
No answers commented yet. Be the first to comment!