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.

Let the first term be \(a = 56\) and the common difference be \(d = 53 - 56 = -3\).

The last term is given as \(-22\). Using the formula for the \(n\)-th term of an arithmetic progression, \(a_n = a + (n-1)d\), we have:

\(-22 = 56 + (n-1)(-3)\)

\(-22 = 56 - 3(n-1)\)

\(-22 = 56 - 3n + 3\)

\(-22 = 59 - 3n\)

\(3n = 59 + 22\)

\(3n = 81\)

\(n = 27\)

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

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

where \(l\) is the last term.

\(S_{27} = \frac{27}{2} (56 + (-22))\)

\(S_{27} = \frac{27}{2} (34)\)

\(S_{27} = 27 \times 17\)

\(S_{27} = 459\)

Given the arithmetic progression with terms \(a, 2a, a^2\), the common difference \(d\) is:

\(d = 2a - a = a\)

\(d = a^2 - 2a\)

Equating the two expressions for \(d\):

\(a = a^2 - 2a\)

\(a^2 - 3a = 0\)

\(a(a - 3) = 0\)

Since \(a\) is positive, \(a = 3\).

Thus, \(d = 3\).

The sum of the first 50 terms \(S_{50}\) is given by:

\(S_{50} = \frac{50}{2} (2a + 49d)\)

Substitute \(a = 3\) and \(d = 3\):

\(S_{50} = 25 (2 \times 3 + 49 \times 3)\)

\(S_{50} = 25 (6 + 147)\)

\(S_{50} = 25 \times 153\)

\(S_{50} = 3825\)

Let the angle of the smallest sector be \(a\) and the common difference be \(d\). The angles of the sectors are \(a, a+d, a+2d, a+3d, a+4d\).

Given that the largest angle is 4 times the smallest angle:

\(a + 4d = 4a\)

\(3a = 4d\)

The sum of the angles in the circle is 360°:

\(360 = S_5 = \frac{5}{2}(2a + 4d)\)

Substitute \(4d = 3a\) into the equation:

\(360 = \frac{5}{2}(2a + 3a) = 12.5a\)

\(a = 28.8^{\circ}\)

The largest angle is \(a + 4d = 4a = 115.2^{\circ}\).

Given the sum of the first \(n\) terms of an arithmetic progression: \(S_n = 32n - n^2\).

To find the first term \(a\), set \(n = 1\):

\(S_1 = 32(1) - 1^2 = 32 - 1 = 31\).

Thus, the first term \(a = 31\).

To find the common difference \(d\), set \(n = 2\):

\(S_2 = 32(2) - 2^2 = 64 - 4 = 60\).

The second term is given by \(S_2 - S_1 = 60 - 31 = 29\).

The common difference \(d = 29 - 31 = -2\).

Therefore, the first term is 31 and the common difference is -2.