(i) For a geometric progression, the ratio between consecutive terms is constant. Therefore, \(\frac{x-3}{x} = \frac{x-5}{x-3}\).

Cross-multiplying gives: \((x-3)^2 = x(x-5)\).

Expanding both sides: \(x^2 - 6x + 9 = x^2 - 5x\).

Simplifying: \(-6x + 9 = -5x\).

Rearranging gives: \(x = 9\).

(ii) The common ratio \(r\) is \(\frac{x-3}{x} = \frac{6}{9} = \frac{2}{3}\).

The fourth term is \(ar^3 = 9 \times \left(\frac{2}{3}\right)^3 = 9 \times \frac{8}{27} = \frac{72}{27} = 2\frac{2}{3}\) or 2.67.

(iii) The sum to infinity \(S_\infty\) is given by \(\frac{a}{1-r}\).

Substituting the values: \(S_\infty = \frac{9}{1 - \frac{2}{3}} = \frac{9}{\frac{1}{3}} = 27\).

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

The sum of the first two terms is given by:

\(a(1 + r) = 15\)

The sum to infinity is given by:

\(\frac{a}{1 - r} = \frac{125}{7}\)

From the first equation, we have:

\(a = \frac{15}{1 + r}\)

Substitute \(a\) in the second equation:

\(\frac{15}{1 + r} \cdot \frac{1}{1 - r} = \frac{125}{7}\)

\(\frac{15}{(1 + r)(1 - r)} = \frac{125}{7}\)

\(\frac{15}{1 - r^2} = \frac{125}{7}\)

\(15 \cdot 7 = 125(1 - r^2)\)

\(105 = 125 - 125r^2\)

\(125r^2 = 20\)

\(r^2 = \frac{4}{25}\)

Since \(r\) is negative, \(r = -\frac{2}{5}\).

Substitute \(r\) back to find \(a\):

\(a = \frac{15}{1 - \frac{2}{5}} = \frac{15}{\frac{3}{5}} = 25\)

The third term is \(ar^2\):

\(ar^2 = 25 \cdot \frac{4}{25} = 4\)

(i) The distances form a geometric sequence with the first term as x and a common ratio of 1.1. The distance on day 21 is given by:

\(x(1.1)^{20} = 20\)

Solving for \(x\):

\(x = \frac{20}{(1.1)^{20}} \approx 3.0\)

Thus, the distance on day 1 is 3.0 km.

(ii) The total distance run over 21 days is the sum of a geometric series:

\(S = x \frac{(1.1)^{21} - 1}{1.1 - 1}\)

Substituting \(x = 3.0\):

\(S = 3.0 \frac{(1.1)^{21} - 1}{0.1} \approx 192\)

Therefore, the total distance run over 21 days is 192 km.

Let the first term be \(a = 12\) and the common ratio be \(r\).

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

The sum of the first four terms \(S_4\) is \(\frac{a(1-r^4)}{1-r}\).

According to the problem, \(S_\infty = 9 \times S_4\).

Substitute the formulas: \(\frac{a}{1-r} = 9 \times \frac{a(1-r^4)}{1-r}\).

Cancel \(a\) and \(1-r\) from both sides: \(1 = 9(1-r^4)\).

Simplify: \(9r^4 = 8\).

Thus, \(r^4 = \frac{8}{9}\).

The fifth term is \(ar^4 = 12 \times \frac{8}{9}\).

Calculate: \(ar^4 = \frac{96}{9} \approx 10.7\).

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

The third term is given by \(ar^2 = 48\).

The fourth term is given by \(ar^3 = 32\).

Divide the second equation by the first to find \(r\):

\(\frac{ar^3}{ar^2} = \frac{32}{48}\)

\(r = \frac{2}{3}\)

Substitute \(r = \frac{2}{3}\) into \(ar^2 = 48\):

\(a \left( \frac{2}{3} \right)^2 = 48\)

\(a \cdot \frac{4}{9} = 48\)

\(a = 108\)

The sum to infinity \(S_\infty\) of a geometric progression is given by:

\(S_\infty = \frac{a}{1 - r}\)

\(S_\infty = \frac{108}{1 - \frac{2}{3}}\)

\(S_\infty = \frac{108}{\frac{1}{3}}\)

\(S_\infty = 324\)