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

The general term of an arithmetic progression is given by \(a + (n-1)d\).

The eighth term is \(a + 7d\) and the third term is \(a + 2d\).

According to the problem, \(a + 7d = 3(a + 2d)\).

Expanding the right side, we have \(a + 7d = 3a + 6d\).

Rearranging gives \(2a = d\).

The sum of the first \(n\) terms of an arithmetic progression is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).

The sum of the first eight terms is \(S_8 = \frac{8}{2} (2a + 7d) = 4(2a + 7d)\).

The sum of the first four terms is \(S_4 = \frac{4}{2} (2a + 3d) = 2(2a + 3d)\).

Substituting \(d = 2a\) into these expressions:

\(S_8 = 4(2a + 7(2a)) = 4(2a + 14a) = 4(16a) = 64a\).

\(S_4 = 2(2a + 3(2a)) = 2(2a + 6a) = 2(8a) = 16a\).

Thus, \(S_8 = 4 \times S_4\).