Let the first term be \(a = 32\) and the common difference be \(d\).

The 5th term is given by \(a + 4d = 22\).

Substituting \(a = 32\), we have:

\(32 + 4d = 22\)

\(4d = 22 - 32\)

\(4d = -10\)

\(d = -2.5\)

The last term is \(-28\), so:

\(a + (n-1)d = -28\)

\(32 + (n-1)(-2.5) = -28\)

\(32 - 2.5(n-1) = -28\)

\(32 - 2.5n + 2.5 = -28\)

\(34.5 - 2.5n = -28\)

\(-2.5n = -28 - 34.5\)

\(-2.5n = -62.5\)

\(n = 25\)

The sum of an arithmetic progression is given by:

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

where \(l\) is the last term.

\(S_{25} = \frac{25}{2} (32 + (-28))\)

\(S_{25} = \frac{25}{2} (4)\)

\(S_{25} = 25 \times 2\)

\(S_{25} = 50\)