(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 \(q = 2\).