Let the first term be \(a = -2222\) and the common difference be \(d = 17\). The \(n\)-th term of an arithmetic progression is given by:

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

We need to find the smallest \(n\) such that \(a_n > 0\).

\(-2222 + (n-1) imes 17 > 0\)

\((n-1) imes 17 > 2222\)

\(n-1 > \frac{2222}{17}\)

\(n-1 > 130.7\)

Since \(n\) must be an integer, \(n = 132\).

Now, calculate the 132nd term:

\(a_{132} = -2222 + 131 imes 17\)

\(a_{132} = -2222 + 2227\)

\(a_{132} = 5\)

Thus, the first positive term is 5.