Letter series B E J Q Z – UGC NET Question

The positions of the letters are B(2), E(5), J(10), Q(17) and Z(26). The differences between consecutive positions are +3, +5, +7 and +9, forming an increasing sequence of odd numbers. The next difference should be +11, so the next position is 26 + 11 = 37, which corresponds to 37 − 26 = 11, the letter K. Thus K uniquely maintains the pattern of increasing odd-numbered jumps with wrap-around.

Option A:
Option A, J, is at position 10, which would move backward from Z instead of forward. This contradicts the established behaviour of positive jumps in position. Since all previous moves are forward and increase in size, returning to J breaks the pattern entirely.

Option B:
Option B, H, lies at position 8, again going backward from 26 to 8. There is no indication of reversal or oscillation in the sequence. Because the pattern involves steadily increasing forward gaps, H cannot be justified as the next letter.

Option C:
Option C is correct because K corresponds to position 11, which is 37 modulo 26, produced by adding 11 to 26. It keeps the next gap as the next odd number after 9 and respects the cyclic nature of the alphabet. This alignment with both the size and direction of the jump makes K the only valid continuation.

Option D:
Option D, M, is at position 13, which would imply a jump of +? that does not match the required +11 step. The position 13 cannot be obtained from 26 by adding any single odd number that fits the observed sequence. Therefore, M does not follow the underlying rule of increasing odd differences.