Which is the (n)th term of the sequence (6,11,18,27,38,\ldots)?

(n^2+2n+3)

The sequence is \\(6,11,18,27,38,\\ldots\\). Its successive differences are 5, 7, 9, and 11, so the difference increases by 2 each time. This suggests a quadratic expression. Test the supplied option \\(a_n=n^2+2n+3\\): for \\(n=1\\), it gives \\(1+2+3=6\\); for \\(n=2\\), it gives \\(4+4+3=11\\); and for \\(n=3\\), it gives \\(9+6+3=18\\). It also gives 27 and 38 for n = 4 and 5. Thus option B is correct.

The formula can be confirmed through the second differences. For a quadratic \\(an^2+bn+c\\), a constant second difference is expected; here it is 2, so the coefficient of \\(n^2\\) is 1. Using the terms then gives the expression \\(n^2+2n+3\\). The nearby choices fail even at the first or second term. Therefore the supplied answer B and its explanation correctly identify the nth term.

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