Which is the (n)th term of the sequence (4,13,28,49,76,\ldots)?

(3n^2+1)

The sequence is 4, 13, 28, 49, 76, and the first differences are 9, 15, 21, and 27. These increase by 6, so the second difference is constant and the nth-term formula is quadratic. Test option B, \\(a_n=3n^2+1\\): for n = 1, it gives \\(3(1)^2+1=4\\); for n = 2, it gives \\(3(2)^2+1=13\\); and for n = 3, it gives \\(3(3)^2+1=28\\). At n = 4 and 5 it gives 49 and 76. Thus option B is correct.

For a quadratic expression \\(an^2+bn+c\\), the second difference is \\(2a\\). Since the second difference here is 6, the coefficient of \\(n^2\\) must be 3. The constant term is then fixed by the first term: \\(3(1)^2+c=4\\), so \\(c=1\\). The formula becomes \\(3n^2+1\\). The other choices do not reproduce the listed terms. Therefore the supplied answer B is mathematically sound.

This is a Class 9 Mathematics question. Chapter: Sequences and Progressions. Topic: nth term.