UGC NET Questions (Paper – 1)

The sequence is generated by aₙ = 3n⁴ + 4n² + 1 for n starting from 1. For n = 1, 2, 3, 4 and 5 this gives 3+4+1 = 8, 48+16+1 = 65, 243+36+1 = 280, 768+64+1 = 833 and 1875+100+1 = 1976, confirming the rule. For n = 6 we compute 3·6⁴+4·6²+1 = 3·1296+144+1 = 3888+144+1 = 4033. Thus 4033 is the only value that maintains this quartic-plus-quadratic-plus-constant pattern.

Option A:
Option A, 3985, is smaller than the formula’s output and does not satisfy aₙ = 3n⁴+4n²+1 for n = 6. It would require subtracting 48 from the correctly computed value only at the final term, which is inconsistent with the earlier behaviour. Therefore option A is incorrect.

Option B:
Option B, 4017, is closer but still not equal to 4033 and again fails to arise from substituting n = 6 into the rule. Choosing 4017 would mean modifying the constant or coefficients in a way not supported by previous terms. Hence option B is incorrect.

Option C:
Option C, 4033, matches exactly the value obtained from the polynomial expression for n = 6. It preserves the relative contributions of the quartic and quadratic terms as well as the constant increment. Because the same rule explains all terms including the next, 4033 is the correct answer.

Option D:
Option D, 4065, overshoots the correct value by 32 and does not equal 3·6⁴+4·6²+1. Using 4065 would break the precise algebraic relationship that defines the series, so option D is incorrect.