(a) The sum to infinity of the geometric progression is given by \(\frac{5a}{1 - \left(-\frac{1}{4}\right)}\).

The sum of the first eight terms of the arithmetic progression is \(\frac{8}{2} [2a + 7(-4)]\).

Equating the two sums: \(\frac{5a}{1 + \frac{1}{4}} = 4a + 8(-4)\).

Simplifying, \(\frac{5a}{\frac{5}{4}} = 8a - 28\).

\(4a = 8a - 112\).

Solving for \(a\), we get \(a = 28\).

(b) The \(k\)th term of the arithmetic progression is given by \(a + (k-1)(-4) = 0\).

Substituting \(a = 28\), we have \(28 + (k-1)(-4) = 0\).

Solving for \(k\), \(28 - 4k + 4 = 0\).

\(32 = 4k\).

\(k = 8\).

(a) For the arithmetic progression, the second term is \(\frac{3}{2}a\) and the third term is \(b\). The common difference \(d\) is given by:

\(d = \frac{3}{2}a - a = \frac{a}{2}\)

\(b = \frac{3}{2}a + \frac{a}{2} = 2a\)

For the geometric progression, the second term is 18 and the third term is \(b + 3\). The common ratio \(r\) is given by:

\(r = \frac{18}{a}\) and \(b + 3 = 18r\)

Substituting \(b = 2a\) into \(b + 3 = 18r\), we get:

\(2a + 3 = 18 \times \frac{18}{a}\)

\(2a^2 + 3a - 324 = 0\)

Solving this quadratic equation, we find \(a = 12\) and \(b = 24\).

(b) The common difference \(d = 6\) (since \(d = \frac{a}{2} = 6\)). The sum of the first 20 terms \(S_{20}\) is given by:

\(S_{20} = \frac{20}{2} \times (2 \times 12 + 19 \times 6)\)

\(S_{20} = 10 \times (24 + 114) = 10 \times 138 = 1380\)

(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.

For Scheme A, the amount of waste recycled each month forms an arithmetic sequence with the first term \(a = 2.5\) and common difference \(d = 0.16\).

The sum of the first \(n\) terms of an arithmetic sequence is given by:

\(S_n = \frac{n}{2} (2a + (n-1)d)\)

Substituting \(n = 24\), \(a = 2.5\), and \(d = 0.16\):

\(S_{24} = \frac{24}{2} (2 \times 2.5 + 23 \times 0.16)\)

\(S_{24} = 12 (5 + 3.68)\)

\(S_{24} = 12 \times 8.68 = 104.16\)

Rounding to the nearest whole number, the total amount is 104 tonnes.

For Scheme B, the amount of waste recycled each month forms a geometric sequence with the first term \(a = 2.5\) and common ratio \(r = 1.06\).

The sum of the first \(n\) terms of a geometric sequence is given by:

\(S_n = a \frac{r^n - 1}{r - 1}\)

Substituting \(n = 24\), \(a = 2.5\), and \(r = 1.06\):

\(S_{24} = 2.5 \frac{1.06^{24} - 1}{1.06 - 1}\)

\(S_{24} = 2.5 \frac{4.29187072 - 1}{0.06}\)

\(S_{24} = 2.5 \times 54.864512\)

\(S_{24} = 137.16128\)

Rounding to the nearest whole number, the total amount is 137 tonnes.

(i) From the arithmetic progression (AP), we have:

\(x - 4 = y - x\)

From the geometric progression (GP), we have:

\(\frac{y}{x} = \frac{18}{y}\)

Solving the GP equation gives:

\(y^2 = 18x\)

Substitute \(x = \frac{y^2}{18}\) into the AP equation:

\(\frac{y^2}{18} - 4 = y - \frac{y^2}{18}\)

Rearrange and simplify:

\(4 + 2\left(\frac{y^2}{18} - 4\right) = y\)

\(y^2 - 9y - 36 = 0\)

Solving this quadratic equation gives:

\(x = 8, y = 12\)

(ii) For the AP, the fourth term is:

\(4 + 3d = 16\)

For the GP, the fourth term is:

\(8 \times \left(\frac{12}{8}\right)^3 = 27\)