(i) The formula for the sum to infinity of a geometric progression is \(\frac{a}{1-r} = 256\), where \(a = 64\). Solving for \(r\):

\(\frac{64}{1-r} = 256\)

\(1-r = \frac{64}{256}\)

\(1-r = \frac{1}{4}\)

\(r = \frac{3}{4}\)

(ii) The formula for the sum of the first \(n\) terms is \(S_n = a \frac{1-r^n}{1-r}\). For the first ten terms:

\(S_{10} = 64 \frac{1-(0.75)^{10}}{1-0.75}\)

\(S_{10} = 64 \frac{1-0.0563}{0.25}\)

\(S_{10} = 64 \times 3.7748\)

\(S_{10} = 242\)

The first term of the geometric progression is given as \(a = 6\).

The second term is given as \(ar = 4\), where \(r\) is the common ratio.

We can find \(r\) by solving the equation \(6r = 4\).

Thus, \(r = \frac{4}{6} = \frac{2}{3}\).

The formula for the sum to infinity of a geometric progression is \(S_\infty = \frac{a}{1-r}\), provided \(|r| < 1\).

Substituting the values, we get \(S_\infty = \frac{6}{1 - \frac{2}{3}}\).

Calculating further, \(S_\infty = \frac{6}{\frac{1}{3}} = 6 \times 3 = 18\).

(i) Let the first term be \(a\) and the common ratio be \(r\). The second term is \(ar = 18\) and the fourth term is \(ar^3 = 8\).

From \(ar = 18\), we have \(a = \frac{18}{r}\).

Substitute \(a\) in \(ar^3 = 8\):

\(\frac{18}{r} \cdot r^3 = 8\)

\(18r^2 = 8\)

\(r^2 = \frac{8}{18} = \frac{4}{9}\)

\(r = \frac{2}{3}\) (since \(r\) is positive)

Substitute \(r = \frac{2}{3}\) back into \(a = \frac{18}{r}\):

\(a = \frac{18}{\frac{2}{3}} = 27\)

(ii) The sum to infinity of a geometric progression is given by \(S = \frac{a}{1-r}\), where \(|r| < 1\).

\(S = \frac{27}{1 - \frac{2}{3}} = \frac{27}{\frac{1}{3}} = 81\)

Let the first term be \(a\) and the common ratio be \(r\).

The third term is given by \(ar^2 = 1764\).

The sum of the second and third terms is \(ar + ar^2 = 3444\).

We have the equations:

1. \(ar^2 = 1764\)

2. \(ar + ar^2 = 3444\)

From equation 1, \(ar^2 = 1764\).

From equation 2, \(ar + ar^2 = 3444\), substitute \(ar^2 = 1764\):

\(ar + 1764 = 3444\)

\(ar = 3444 - 1764 = 1680\)

Now, divide the two equations:

\(\frac{ar^2}{ar} = \frac{1764}{1680}\)

\(r = \frac{1764}{1680} = \frac{21}{20} = 1.05\)

Substitute \(r = 1.05\) back into \(ar = 1680\):

\(a \times 1.05 = 1680\)

\(a = \frac{1680}{1.05} = 1600\)

The 50th term is given by \(ar^{49}\):

\(1600 \times (1.05)^{49} \approx 17,500\)

(a) The sequence is a geometric progression with first term \(a = 50\) and common ratio \(r = \frac{40}{50} = 0.8\).

The general term is given by \(a_n = 50 \times 0.8^{n-1}\).

We need to find the smallest \(n\) such that \(a_n < 10\).

\(50 \times 0.8^{n-1} < 10\)

\(0.8^{n-1} < 0.2\)

Using logarithms, \((n-1) \log(0.8) < \log(0.2)\)

\(n-1 > \frac{\log(0.2)}{\log(0.8)} \approx 7.2\)

\(n > 8.2\), so \(n = 9\).

Hence, the 9th impact is required.

(b) The sum of the first 20 terms of a geometric progression is given by:

\(S_{20} = \frac{50(1 - 0.8^{20})}{1 - 0.8}\)

\(S_{20} = \frac{50(1 - 0.8^{20})}{0.2}\)

\(S_{20} = 50 \times 5 \times (1 - 0.8^{20})\)

\(S_{20} \approx 247\) mm (to the nearest millimetre).

(c) The greatest total depth is the sum to infinity of the geometric progression:

\(S_\infty = \frac{50}{1 - 0.8}\)

\(S_\infty = \frac{50}{0.2}\)

\(S_\infty = 250\) mm.