In the number series 4, 14, 33, 72, 195, the next term is _______. – UGC NET Question

This sequence is modelled by the rule aₙ = n! + 3n² with n starting from 1. For n = 1, 2, 3, 4 and 5 the expression gives 1! + 3·1² = 4, 2! + 3·2² = 14, 3! + 3·3² = 33, 4! + 3·4² = 72 and 5! + 3·5² = 195, which matches the given terms exactly. For n = 6 we obtain a₆ = 6! + 3·6² = 720 + 108 = 828. Hence 828 is the only value consistent with this factorial-based pattern.

Option A:
Option A, 804, is 24 less than the value from n! + 3n² at n = 6 and cannot be generated by the same formula. Choosing 804 would require subtracting 24 from the correct result only at the sixth term, which disrupts the pattern of factorial growth with a quadratic adjustment. Therefore option A is incorrect.

Option B:
Option B, 816, is 12 below the correct value 828 and does not equal 720 + 108. It represents a near approximation but still violates the exact functional relationship between n and aₙ. Consequently, option B does not correctly continue the series.

Option C:
Option C, 840, is 12 greater than the required value and likewise cannot arise from 6! + 3·6². Accepting 840 would artificially inflate the last term beyond what the rule predicts. Thus option C is not a valid continuation of the sequence.

Option D:
Option D, 828, exactly matches the value produced by the rule when n = 6. It faithfully maintains the combination of factorial growth and quadratic adjustment that governs all previous terms. Because it extends this pattern without any exception, option D is the correct answer.