n + 1
2n + 1
3n + 1
4n + 1
Correct answer is B
\(S_{n} = \frac{n}{2}(2a + (n - 1) d = n^{2} + 2n\)
\(n(2a + (n - 1) d = 2n^{2} + 4n\)
\(2an + n^{2}d - nd = 2n^{2} + 4n\)
\(n^{2}d = 2n^{2}\)
\(d = 2\)
\((2a - d) n = 4n\)
\(2a - d = 4 \implies 2a = 4 + d = 4 + 2 = 6\)
\(a = 3\)
\(T_{n} = a + (n - 1)d\)
= \(3 + (n - 1)2 = 3 + 2n - 2 = 2n + 1\)