The first term \(a = 6\) and the second term is \(\frac{6^2}{6+2} = \frac{36}{8} = \frac{9}{2}\).

The common difference \(d\) is given by:

\(d = \frac{9}{2} - 6 = \frac{9}{2} - \frac{12}{2} = -\frac{3}{2}\)

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

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

Substituting the values, we have:

\(S_n = \frac{n}{2} \left( 12 + (n-1)(-\frac{3}{2}) \right)\)

We need \(S_n < -480\):

\(\frac{n}{2} \left( 12 - \frac{3}{2}(n-1) \right) < -480\)

\(n \left( 12 - \frac{3}{2}n + \frac{3}{2} \right) < -960\)

\(n \left( \frac{27}{2} - \frac{3}{2}n \right) < -960\)

\(27n - 3n^2 < -1920\)

\(3n^2 - 27n - 1920 > 0\)

Solving the quadratic inequality:

\(n = \frac{9 \pm \sqrt{81 + 2560}}{2}\)

\(n = \frac{9 \pm \sqrt{2641}}{2}\)

Calculating the roots, we find the least integer \(n\) is 31.