(i) Boxer A's weight loss follows an arithmetic progression with first term \(a = 1\) kg and second term \(0.98\) kg. The common difference \(d = 0.98 - 1 = -0.02\) kg. The total weight loss after \(x\) weeks is given by the sum of an arithmetic series:

\(S_x = \frac{x}{2} [2a + (x-1)d] = \frac{x}{2} [2 + (x-1)(-0.02)]\)

(ii) To find \(n\) when the total weight loss is 13 kg, set \(S_n = 13\):

\(\frac{n}{2} [2 + (n-1)(-0.02)] = 13\)

Simplifying gives a quadratic equation, which can be solved to find \(n = 16\).

(iii) Boxer B's weight loss follows a geometric progression with first term \(a = 1\) kg and common ratio \(r = 0.92\). The total weight loss after 20 weeks is given by:

\(S_{20} = \frac{a(1-r^{20})}{1-r} = \frac{1(1-0.92^{20})}{1-0.92} \approx 10.1 \text{ kg}\)

The sum to infinity \(S_\infty = \frac{a}{1-r} = \frac{1}{0.08} = 12.5 \text{ kg}\), so Boxer B can never reach the 13 kg target.