The sum of the first \(n\) terms of an arithmetic progression is given by:

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

Given \(S_n = 2n^2 + 8n\), we equate this to the formula:

\(\frac{n}{2} (2a + (n-1)d) = 2n^2 + 8n\)

For \(n = 1\):

\(S_1 = 2(1)^2 + 8(1) = 10\)

Thus, \(a = 10\).

For \(n = 2\):

\(S_2 = 2(2)^2 + 8(2) = 24\)

\(S_2 = a + (a + d) = 24\)

\(10 + (10 + d) = 24\)

\(20 + d = 24\)

\(d = 4\)

Therefore, the first term is 10 and the common difference is 4.