Given the sum of the first n terms \(S_n = \frac{1}{2}n(3n + 7)\).

For an arithmetic progression, the sum of the first n terms is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Equating the two expressions for \(S_n\):

\(\frac{n}{2} (2a + (n-1)d) = \frac{1}{2}n(3n + 7)\)

Cancel \(\frac{n}{2}\) from both sides:

\(2a + (n-1)d = 3n + 7\)

Substitute \(n = 1\):

\(2a = 3(1) + 7\)

\(2a = 10\)

\(a = 5\)

Substitute \(n = 2\):

\(2a + d = 3(2) + 7\)

\(2(5) + d = 13\)

\(10 + d = 13\)

\(d = 3\)

Thus, the first term is 5 and the common difference is 3.