The first term of the arithmetic progression is a = p and the second term is a + d = 2p, so the common difference d = p.

The nth term of an arithmetic progression is given by:

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

Substituting the known values:

\(p + (n-1)p = 336\)

\(np = 336\)

The sum of the first n terms is given by:

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

Substituting the known values:

\(\frac{n}{2} (2p + (n-1)p) = 7224\)

\(\frac{n}{2} (p + np) = 7224\)

Substituting \(np = 336\) into the sum equation:

\(\frac{n}{2} (p + 336) = 7224\)

Solving for n and p:

\(n(p + 336) = 14448\)

Using \(np = 336\), we have:

\(n = \frac{336}{p}\)

Substitute into the sum equation:

\(\frac{336}{p} (p + 336) = 14448\)

Solving gives:

\(336 + \frac{336^2}{p} = 14448\)

\(\frac{336^2}{p} = 14448 - 336\)

\(\frac{336^2}{p} = 14112\)

\(p = \frac{336^2}{14112}\)

\(p = 8\)

Substitute back to find n:

\(n = \frac{336}{8} = 42\)

\(Thus, the values are n = 42 and p = 8.\)