The nth term of the arithmetic progression is given by \(a_n = \frac{1}{2}(3n - 15)\).

For the first term \(n = 1\), \(a_1 = \frac{1}{2}(3 \times 1 - 15) = -6\).

For the second term \(n = 2\), \(a_2 = \frac{1}{2}(3 \times 2 - 15) = -4.5\).

The first term \(a = -6\) and the common difference \(d = -4.5 - (-6) = 1.5\).

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

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

Substitute \(S_n = 84\), \(a = -6\), and \(d = 1.5\):

\(84 = \frac{n}{2} (2(-6) + (n-1)1.5)\)

\(84 = \frac{n}{2} (-12 + 1.5n - 1.5)\)

\(84 = \frac{n}{2} (1.5n - 13.5)\)

\(168 = n(1.5n - 13.5)\)

\(168 = 1.5n^2 - 13.5n\)

\(1.5n^2 - 13.5n - 168 = 0\)

Divide the entire equation by 1.5:

\(n^2 - 9n - 112 = 0\)

Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -9\), \(c = -112\):

\(n = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \times 1 \times (-112)}}{2 \times 1}\)

\(n = \frac{9 \pm \sqrt{81 + 448}}{2}\)

\(n = \frac{9 \pm \sqrt{529}}{2}\)

\(n = \frac{9 \pm 23}{2}\)

\(n = 16\) or \(n = -7\)

Since n must be positive, \(n = 16\).