(i) The sum to infinity of a geometric progression is given by \(S = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.

For progression \(P\), \(a = 2\) and \(r = \frac{1}{2}\), so \(S_P = \frac{2}{1 - \frac{1}{2}} = 4\).

For progression \(Q\), \(a = 3\) and \(r = \frac{1}{3}\), so \(S_Q = \frac{3}{1 - \frac{1}{3}} = \frac{9}{2}\).

The sums \(S_P, S_Q, S_R\) form an arithmetic progression, so \(S_R = 5\).

(ii) Given the first term of \(R\) is 4, and \(S_R = \frac{4}{1-r} = 5\), solving for \(r\) gives \(r = \frac{1}{5}\).

The sum of the first three terms of \(R\) is \(4 + 4 \times \frac{1}{5} + 4 \times \left(\frac{1}{5}\right)^2 = 4 + \frac{4}{5} + \frac{4}{25} = \frac{24}{5}\) or 4.96.