(i) Let the common difference be \(d\). The sum of the first ten terms \(S_{10}\) is given by:

\(S_{10} = \frac{10}{2} (2a + 9d) = 5(2a + 9d)\)

The sum of the next five terms \(S_{11 \text{ to } 15}\) is:

\(S_{11 \text{ to } 15} = \frac{5}{2} [(a + 10d) + (a + 14d)] = 5(a + 12d)\)

Given \(S_{10} = S_{11 \text{ to } 15}\), we have:

\(5(2a + 9d) = 5(a + 12d)\)

\(2a + 9d = a + 12d\)

\(a = 3d\)

Thus, \(d = \frac{a}{3}\).

(ii) The tenth term \(a + 9d\) is 36 more than the fourth term \(a + 3d\):

\(a + 9d = (a + 3d) + 36\)

\(6d = 36\)

\(d = 6\)

Since \(d = \frac{a}{3}\), we have:

\(\frac{a}{3} = 6\)

\(a = 18\)

The first term of the arithmetic progression is a = p and the second term is a + d = 2p, so the common difference d = p.

The nth term of an arithmetic progression is given by:

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

Substituting the known values:

\(p + (n-1)p = 336\)

\(np = 336\)

The sum of the first n terms is given by:

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

Substituting the known values:

\(\frac{n}{2} (2p + (n-1)p) = 7224\)

\(\frac{n}{2} (p + np) = 7224\)

Substituting \(np = 336\) into the sum equation:

\(\frac{n}{2} (p + 336) = 7224\)

Solving for n and p:

\(n(p + 336) = 14448\)

Using \(np = 336\), we have:

\(n = \frac{336}{p}\)

Substitute into the sum equation:

\(\frac{336}{p} (p + 336) = 14448\)

Solving gives:

\(336 + \frac{336^2}{p} = 14448\)

\(\frac{336^2}{p} = 14448 - 336\)

\(\frac{336^2}{p} = 14112\)

\(p = \frac{336^2}{14112}\)

\(p = 8\)

Substitute back to find n:

\(n = \frac{336}{8} = 42\)

\(Thus, the values are n = 42 and p = 8.\)

The formula for the nth term of an arithmetic progression is given by:

\(a + (n-1) imes 3 = 94\)

For the sum of the first n terms, we use:

\(\frac{n}{2} [2a + (n-1) imes 3] = 1420\)

Substituting the first equation into the second:

\(\frac{n}{2} [a + 94] = 1420\)

Attempt elimination of a or n:

\(3n^2 - 191n + 2840 = 0\)

Solving this quadratic equation gives:

\(n = 40\)

Substitute back to find a:

\(a + 39 imes 3 = 94\)

\(a + 117 = 94\)

\(a = -23\)

The first term of the arithmetic progression is given as \(a = 6\).

The second term is \(2\), so the common difference \(d\) is \(2 - 6 = -4\).

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

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

Substitute \(n = 80\), \(a = 6\), and \(d = -4\) into the formula:

\(S_{80} = \frac{80}{2} (2 \times 6 + 79 \times (-4))\)

\(S_{80} = 40 (12 - 316)\)

\(S_{80} = 40 \times (-304)\)

\(S_{80} = -12,160\)

(i) The first term of the progression is \(p + q\) and the last term is \(p + qn\). The sum of the first n terms, \(S_n\), is given by the formula for the sum of an arithmetic series:

\(S_n = \frac{n}{2} \left( 2(p + q) + (n-1)q \right)\)

or equivalently,

\(S_n = \frac{n}{2} (2p + q + nq)\)

(ii) Using the expression for \(S_n\) from part (i), we have:

\(2(2p + q + 4q) = 40\)

\(3(2p + q + 6q) = 72\)

Solving these equations simultaneously:

From the first equation: \(2(2p + 5q) = 40\) gives \(2p + 5q = 20\).

From the second equation: \(3(2p + 7q) = 72\) gives \(2p + 7q = 24\).

Subtract the first equation from the second:

\((2p + 7q) - (2p + 5q) = 24 - 20\)

\(2q = 4\)

\(q = 2\)

Substitute \(q = 2\) back into \(2p + 5q = 20\):

\(2p + 5(2) = 20\)

\(2p + 10 = 20\)

\(2p = 10\)

\(p = 5\)

Thus, \(p = 5\) and \(q = 2\).