The first term \(a = 6\) and the second term is \(\frac{6^2}{6+2} = \frac{36}{8} = \frac{9}{2}\).

The common difference \(d\) is given by:

\(d = \frac{9}{2} - 6 = \frac{9}{2} - \frac{12}{2} = -\frac{3}{2}\)

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

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

Substituting the values, we have:

\(S_n = \frac{n}{2} \left( 12 + (n-1)(-\frac{3}{2}) \right)\)

We need \(S_n < -480\):

\(\frac{n}{2} \left( 12 - \frac{3}{2}(n-1) \right) < -480\)

\(n \left( 12 - \frac{3}{2}n + \frac{3}{2} \right) < -960\)

\(n \left( \frac{27}{2} - \frac{3}{2}n \right) < -960\)

\(27n - 3n^2 < -1920\)

\(3n^2 - 27n - 1920 > 0\)

Solving the quadratic inequality:

\(n = \frac{9 \pm \sqrt{81 + 2560}}{2}\)

\(n = \frac{9 \pm \sqrt{2641}}{2}\)

Calculating the roots, we find the least integer \(n\) is 31.

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

Equating this to \(n^2 + 4n\), we have:

\(\frac{n}{2} (2a + (n-1)d) = n^2 + 4n\).

Comparing coefficients, we get:

\(\frac{d}{2} = 1\) and \(a - \frac{1}{2}d = 4\).

Solving these, \(d = 2\) and \(a = 5\).

The \(k\)th term is given by \(a + (k-1)d\).

Substituting the values of \(a\) and \(d\), we have:

\(5 + 2(k-1) > 200\).

Simplifying, \(2k + 3 > 200\).

Solving for \(k\), we get \(k > 98.5\).

Thus, the smallest integer value of \(k\) is \(99\).

The nth term of the arithmetic progression is given by \(a_n = \frac{1}{2}(3n - 15)\).

For the first term \(n = 1\), \(a_1 = \frac{1}{2}(3 \times 1 - 15) = -6\).

For the second term \(n = 2\), \(a_2 = \frac{1}{2}(3 \times 2 - 15) = -4.5\).

The first term \(a = -6\) and the common difference \(d = -4.5 - (-6) = 1.5\).

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

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

Substitute \(S_n = 84\), \(a = -6\), and \(d = 1.5\):

\(84 = \frac{n}{2} (2(-6) + (n-1)1.5)\)

\(84 = \frac{n}{2} (-12 + 1.5n - 1.5)\)

\(84 = \frac{n}{2} (1.5n - 13.5)\)

\(168 = n(1.5n - 13.5)\)

\(168 = 1.5n^2 - 13.5n\)

\(1.5n^2 - 13.5n - 168 = 0\)

Divide the entire equation by 1.5:

\(n^2 - 9n - 112 = 0\)

Using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -9\), \(c = -112\):

\(n = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \times 1 \times (-112)}}{2 \times 1}\)

\(n = \frac{9 \pm \sqrt{81 + 448}}{2}\)

\(n = \frac{9 \pm \sqrt{529}}{2}\)

\(n = \frac{9 \pm 23}{2}\)

\(n = 16\) or \(n = -7\)

Since n must be positive, \(n = 16\).

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

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

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

For the first nine terms:

\(S_9 = \frac{9}{2} (2a + 8d) = 117\)

Solving for \(2a + 8d\):

\(9(2a + 8d) = 234\)

\(2a + 8d = 26\) (Equation 1)

The sum of the next four terms (terms 10 to 13) is 91:

\(S_{13} - S_9 = 91\)

\(S_{13} = \frac{13}{2} (2a + 12d)\)

\(\frac{13}{2} (2a + 12d) - \frac{9}{2} (2a + 8d) = 91\)

\(13(2a + 12d) - 9(2a + 8d) = 182\)

\(26a + 156d - 18a - 72d = 182\)

\(8a + 84d = 182\)

\(4a + 42d = 91\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 1: \(2a + 8d = 26\)

From Equation 2: \(4a + 42d = 91\)

Multiply Equation 1 by 2:

\(4a + 16d = 52\)

Subtract from Equation 2:

\((4a + 42d) - (4a + 16d) = 91 - 52\)

\(26d = 39\)

\(d = 1.5\)

Substitute \(d = 1.5\) into Equation 1:

\(2a + 8(1.5) = 26\)

\(2a + 12 = 26\)

\(2a = 14\)

\(a = 7\)

Thus, the first term is 7, and the common difference is 1.5.

(i) The distance run on the last day is the 21st term of an arithmetic sequence where the first term is 13 km and the common difference is 1.2 km. The formula for the nth term of an arithmetic sequence is:

\(a_n = a + (n-1) imes d\)

Substituting the given values:

\(a_{21} = 13 + 20 imes 1.2 = 37\) km

(ii) The total distance run over 21 days is the sum of the arithmetic sequence. The formula for the sum of the first n terms is:

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

where \(l\) is the last term. Substituting the given values:

\(S_{21} = \frac{21}{2} \times (13 + 37) = 525\) km