(i) Let the first term be \(a = 81\) and the common ratio be \(r\). The fourth term is given by \(ar^3 = 24\).

Thus, \(81r^3 = 24\) which gives \(r^3 = \frac{24}{81}\).

Solving for \(r\), we get \(r = \frac{2}{3}\) or 0.667.

(ii) The sum to infinity of a geometric progression is given by \(S_\infty = \frac{a}{1-r}\).

Substituting \(a = 81\) and \(r = \frac{2}{3}\), we get \(S_\infty = \frac{81}{1 - \frac{2}{3}} = 243\).

(iii) The second term of the GP is \(ar = 81 \times \frac{2}{3} = 54\).

The third term of the GP is \(ar^2 = 36\).

These are the first and fourth terms of the AP, so \(3d = -18\) giving \(d = -6\).

The sum of the first ten terms of the AP is \(S_{10} = 5 \times (108 - 54) = 270\).