Three-letter groups step of three cyclic – UGC NET Question

The starting letters M, P, S and V occupy positions 13, 16, 19 and 22, increasing by 3 each time. Each group is formed by three consecutive letters in forward order. Adding 3 to 22 gives 25, which is Y, and the next two letters after Y are Z and A, giving YZA with cyclic wrap around. Thus YZA is the unique group that continues the plus three pattern while respecting the cyclic nature of the alphabet.

Option A:
Option A, XYZ, starts at X, position 24, which is only 2 positions after V instead of 3. It therefore does not follow the consistent step of three seen in the starting letters. Even though XYZ is a consecutive group, it does not arise from the same generating rule.

Option B:
Option B, WXY, begins at W, position 23, just one step after V. This shrinks the required step and contradicts the established difference of three positions between starting letters. For this reason, WXY cannot be the correct continuation.

Option C:
Option C is correct because Y is position 25, exactly three positions beyond V, and YZA forms a three letter block progressing alphabetically with wrap around to A. It respects both the distance between starting letters and the internal consecutive pattern. This makes YZA the logical next term in the series.

Option D:
Option D, ZAB, starts at Z, which is four positions ahead of V when counting cyclically, not three. That would increase the gap and break the regularity of the step size. As the pattern does not show any such change, ZAB cannot be accepted.