Let the first term be \(a = -10\) and the common difference be \(d\).

The 15th term is given by \(a + 14d = 11\).

Substituting \(a = -10\), we have:

\(-10 + 14d = 11\)

\(14d = 21\)

\(d = \frac{3}{2}\)

The last term is given by \(a + (n-1)d = 41\).

Substituting \(a = -10\) and \(d = \frac{3}{2}\), we have:

\(-10 + (n-1) \times \frac{3}{2} = 41\)

\((n-1) \times \frac{3}{2} = 51\)

\(n-1 = 34\)

\(n = 35\)

The sum of the arithmetic progression is given by:

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

\(S_{35} = \frac{35}{2} (-10 + 41)\)

\(S_{35} = \frac{35}{2} \times 31\)

\(S_{35} = 542.5\)