Let the first term be \(a = 12\) and the common difference be \(d\).

The sum of the first four terms is given by:

\(S_4 = \frac{4}{2} (2a + 3d) = 57\)

\(2(24 + 3d) = 57\)

\(48 + 6d = 57\)

\(6d = 9\)

\(d = 1.5\)

The last term \(l = 48\) is given by:

\(l = a + (n-1)d\)

\(48 = 12 + (n-1)1.5\)

\(48 = 12 + 1.5n - 1.5\)

\(48 = 10.5 + 1.5n\)

\(37.5 = 1.5n\)

\(n = 25\)