Given the sum of the first n terms of an arithmetic progression is \(S_n = n^2 + 8n\).

For the first term \(S_1\):

\(S_1 = 1^2 + 8 \times 1 = 9\)

Thus, the first term \(a = 9\).

For the second term \(S_2\):

\(S_2 = 2^2 + 8 \times 2 = 20\)

We know \(S_2 = a + (a + d)\), so:

\(20 = 9 + (9 + d)\)

\(20 = 18 + d\)

\(d = 2\)

Therefore, the first term is 9 and the common difference is 2.

(i) The sum of an arithmetic progression is given by the formula:

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

Given \(S_n = 525\), \(n = 25\), and \(a = -15\), we have:

\(525 = \frac{25}{2} (2(-15) + 24d)\)

\(525 = \frac{25}{2} (-30 + 24d)\)

\(525 = 25(-15 + 12d)\)

\(21 = -15 + 12d\)

\(36 = 12d\)

\(d = 3\)

(ii) The last term \(l\) in the progression is given by:

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

\(l = -15 + 24 \times 3\)

\(l = -15 + 72\)

\(l = 57\)

(iii) The positive terms start from 3 and go up to 57. The first positive term is 3, which is the second term in the sequence:

\(a + d = 3\)

\(-15 + 3 = 3\)

The number of positive terms is 19 (from 3 to 57). The sum of these terms is:

\(S_{19} = \frac{19}{2} (3 + 57)\)

\(S_{19} = \frac{19}{2} \times 60\)

\(S_{19} = 570\)

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

The sixth term is given by \(a + 5d = 23\).

The sum of the first ten terms is given by \(\frac{10}{2} (2a + 9d) = 200\), simplifying to \(5(2a + 9d) = 200\).

Solving the equations:

1. \(a + 5d = 23\)

2. \(5(2a + 9d) = 200\) simplifies to \(2a + 9d = 40\).

From equation 1, \(a = 23 - 5d\).

Substitute \(a = 23 - 5d\) into equation 2:

\(2(23 - 5d) + 9d = 40\)

\(46 - 10d + 9d = 40\)

\(46 - d = 40\)

\(d = 6\)

Substitute \(d = 6\) back into \(a = 23 - 5d\):

\(a = 23 - 5(6)\)

\(a = 23 - 30\)

\(a = -7\)

The seventh term is \(a + 6d = -7 + 6(6) = -7 + 36 = 29\).

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

The general term of an arithmetic progression is given by \(a + (n-1)d\).

The eighth term is \(a + 7d\) and the third term is \(a + 2d\).

According to the problem, \(a + 7d = 3(a + 2d)\).

Expanding the right side, we have \(a + 7d = 3a + 6d\).

Rearranging gives \(2a = d\).

The sum of the first \(n\) terms of an arithmetic progression is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).

The sum of the first eight terms is \(S_8 = \frac{8}{2} (2a + 7d) = 4(2a + 7d)\).

The sum of the first four terms is \(S_4 = \frac{4}{2} (2a + 3d) = 2(2a + 3d)\).

Substituting \(d = 2a\) into these expressions:

\(S_8 = 4(2a + 7(2a)) = 4(2a + 14a) = 4(16a) = 64a\).

\(S_4 = 2(2a + 3(2a)) = 2(2a + 6a) = 2(8a) = 16a\).

Thus, \(S_8 = 4 \times S_4\).

Let the angle of the smallest sector be \(a\) and the common difference of the arithmetic progression be \(d\). The angles of the sectors are \(a, a+d, a+2d, a+3d, a+4d, a+5d\).

Given that the largest angle is 4 times the smallest angle, we have:

\(a + 5d = 4a\)

\(5d = 3a\)

The sum of the angles in the circle is \(360^{\circ}\), so:

\(6a + 15d = 360\)

Substitute \(d = \frac{3a}{5}\) into the equation:

\(6a + 15\left(\frac{3a}{5}\right) = 360\)

\(6a + 9a = 360\)

\(15a = 360\)

\(a = 24^{\circ}\)

The arc length of the smallest sector is:

\(\frac{24}{360} \times 2\pi \times 5 = \frac{2\pi}{15} \times 5 = \frac{2\pi}{3}\)

The perimeter of the smallest sector is the sum of the arc length and the two radii:

\(\frac{2\pi}{3} + 2 \times 5 = \frac{2\pi}{3} + 10\)

Approximating \(\pi \approx 3.1416\), the perimeter is:

\(\frac{2 \times 3.1416}{3} + 10 \approx 12.1 \text{ cm}\)