(a)(i) For an arithmetic progression, the difference between consecutive terms is constant. Thus, \(k^2 - 4k = 8k - k^2\). Solving gives:

\(2k^2 = 12k\)

\(k(k - 6) = 0\)

Since \(k\) is non-zero, \(k = 6\).

(a)(ii) The first term \(a = 4k = 24\) and the common difference \(d = k^2 - 4k = 12\). The sum of the first 20 terms is:

\(S_{20} = \frac{20}{2} (2 \times 24 + 19 \times 12) = 2760\)

(b) Let the first term be \(a\) and the common ratio be \(r\). Then \(ar^3 = 36\) and \(ar^5 = 6\). Dividing gives:

\(r^2 = \frac{1}{6}\)

\(r = \frac{\sqrt{6}}{6}\)

\(a = \frac{36}{r^3} = \frac{1296}{\sqrt{6}}\)

The sum to infinity is:

\(S_\infty = \frac{a}{1 - r} = \frac{1296}{\sqrt{6} - 1}\)