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

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

Given \(S_n = 2n^2 + 8n\), we equate this to the formula:

\(\frac{n}{2} (2a + (n-1)d) = 2n^2 + 8n\)

For \(n = 1\):

\(S_1 = 2(1)^2 + 8(1) = 10\)

Thus, \(a = 10\).

For \(n = 2\):

\(S_2 = 2(2)^2 + 8(2) = 24\)

\(S_2 = a + (a + d) = 24\)

\(10 + (10 + d) = 24\)

\(20 + d = 24\)

\(d = 4\)

Therefore, the first term is 10 and the common difference is 4.

Let the first term be \(a = 12\) and the common difference be \(d\).

The sum of the first four terms is given by:

\(S_4 = \frac{4}{2} (2a + 3d) = 57\)

\(2(24 + 3d) = 57\)

\(48 + 6d = 57\)

\(6d = 9\)

\(d = 1.5\)

The last term \(l = 48\) is given by:

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

\(48 = 12 + (n-1)1.5\)

\(48 = 12 + 1.5n - 1.5\)

\(48 = 10.5 + 1.5n\)

\(37.5 = 1.5n\)

\(n = 25\)

Let the first term of the arithmetic progression be \(a = 3\) and the common difference be \(d = 2\).

The angles of the sectors form an arithmetic sequence: \(3, 5, 7, \ldots\).

The sum of the angles in a circle is 360°.

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

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

Substitute \(a = 3\), \(d = 2\), and \(S_n = 360\):

\(\frac{n}{2} (6 + (n-1)2) = 360\)

Simplify:

\(\frac{n}{2} (6 + 2n - 2) = 360\)

\(\frac{n}{2} (2n + 4) = 360\)

\(n(n + 2) = 360\)

\(n^2 + 2n - 360 = 0\)

Solving the quadratic equation \(n^2 + 2n - 360 = 0\) using the quadratic formula:

\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1\), \(b = 2\), \(c = -360\).

\(n = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-360)}}{2 \times 1}\)

\(n = \frac{-2 \pm \sqrt{4 + 1440}}{2}\)

\(n = \frac{-2 \pm \sqrt{1444}}{2}\)

\(n = \frac{-2 \pm 38}{2}\)

\(n = 18 \text{ or } n = -20\)

Since \(n\) must be positive, \(n = 18\).

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

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

Substituting the given values, \(a = 4\) and \(S_n = 5863\):

\(\frac{n}{2} [8 + (n-1)d] = 5863\)

Multiply both sides by 2:

\(n[8 + (n-1)d] = 11726\)

Divide by \(n\):

\(8 + (n-1)d = \frac{11726}{n}\)

Rearrange to find \((n-1)d\):

\((n-1)d = \frac{11726}{n} - 8\)

(b) The \(n\)th term \(u_n\) is given by:

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

Given \(u_n = 139\):

\(4 + (n-1)d = 139\)

Substitute \((n-1)d = \frac{11726}{n} - 8\):

\(\frac{11726}{n} - 8 = 135\)

\(\frac{11726}{n} = 143\)

\(n = \frac{11726}{143} = 82\)

Substitute \(n = 82\) back into \((n-1)d = \frac{11726}{n} - 8\):

\(81d = \frac{11726}{82} - 8\)

\(81d = 143 - 8\)

\(81d = 135\)

\(d = \frac{135}{81} = \frac{5}{3}\)

The first term of the arithmetic progression is given as 61, so \(a = 61\).

The second term is 57, so the common difference \(d\) is \(57 - 61 = -4\).

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

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

Given that the sum \(S_n = n\), we have:

\(n = \frac{n}{2} [2 \times 61 + (n-1)(-4)]\)

Simplifying inside the brackets:

\(n = \frac{n}{2} [122 + (n-1)(-4)]\)

\(n = \frac{n}{2} [122 - 4n + 4]\)

\(n = \frac{n}{2} [126 - 4n]\)

Multiply both sides by 2 to eliminate the fraction:

\(2n = n(126 - 4n)\)

\(2n = 126n - 4n^2\)

Rearrange to form a quadratic equation:

\(4n^2 - 124n = 0\)

Factor out \(n\):

\(n(4n - 124) = 0\)

So, \(n = 0\) or \(4n - 124 = 0\).

Since \(n\) must be positive, solve \(4n - 124 = 0\):

\(4n = 124\)

\(n = 31\)

Thus, the value of the positive integer \(n\) is 31.