The first term of the arithmetic progression is given as \(a = 5\), and the second term is \(9\), so the common difference \(d = 9 - 5 = 4\).

The general formula for the \(n\)-th term of an arithmetic progression is \(a_n = a + (n-1)d\).

We need to find the smallest \(n\) such that \(a_n > 200\).

Set \(a_n = 200\):

\(200 = 5 + (n-1) imes 4\)

\(200 = 5 + 4n - 4\)

\(200 = 4n + 1\)

\(199 = 4n\)

\(n = \frac{199}{4} = 49.75\)

Since \(n\) must be an integer, we take \(n = 50\) as the smallest integer greater than 49.75.

Now, calculate the sum of the first 50 terms using the sum formula for an arithmetic progression:

\(S_n = \frac{n}{2} (a + a_n)\)

\(S_{50} = \frac{50}{2} (5 + 201)\)

\(S_{50} = 25 imes 206\)

\(S_{50} = 5150\)