(a) The common difference \(d\) of the arithmetic progression is given by:

\(d = (2p - 6) - \frac{p^2}{6} = p - (2p - 6)\)

Equating the two expressions for \(d\):

\(2(2p - 6) = p + \frac{p^2}{6}\)

\(\Rightarrow p^2 - 18p + 72 = 0\)

Solving this quadratic equation by factorization:

\((p - 6)(p - 12) = 0\)

Since \(d \neq 0\), \(p = 12\).

(b) For the geometric progression, the common ratio \(r\) is:

\(r = \frac{2p - 6}{\frac{p^2}{6}} = \frac{18}{24} = \frac{3}{4}\)

The sum to infinity \(S\) is given by:

\(S = \frac{24}{1 - \frac{3}{4}} = 96\)

Let the first term of both progressions be 3. For the arithmetic progression (AP), let the common difference be \(d\). The terms are:

1st term: \(3\)

3rd term: \(3 + 2d\)

13th term: \(3 + 12d\)

For the geometric progression (GP), let the common ratio be \(r\). The terms are:

1st term: \(3\)

2nd term: \(3r\)

3rd term: \(3r^2\)

Equating the terms of the AP and GP:

\(3 + 2d = 3r\)

\(3 + 12d = 3r^2\)

From \(3 + 2d = 3r\), we get \(r = \frac{3 + 2d}{3}\).

Substitute \(r\) in the second equation:

\(3 + 12d = 3\left(\frac{3 + 2d}{3}\right)^2\)

\((3 + 2d)^2 = 3(3 + 12d)\)

\(9 + 12d + 4d^2 = 9 + 36d\)

\(4d^2 - 24d = 0\)

\(4d(d - 6) = 0\)

\(d = 0\) or \(d = 6\)

Since \(d = 0\) is not valid, \(d = 6\).

Substitute \(d = 6\) back to find \(r\):

\(r = \frac{3 + 2(6)}{3} = \frac{15}{3} = 5\)

Thus, the common difference is 6 and the common ratio is 5.

(i) (a) Using model A, the sequence of heights is arithmetic with the first term 0.96 m and common difference -0.04 m. The total distance is twice the sum of the first 20 terms plus the initial drop of 1 m:

\(S_{20} = \frac{20}{2} \times (2 \times 0.96 + 19 \times (-0.04)) = 19.2 - 0.76 = 18.44 \text{ m}\)

\(Total distance = 1 + 2 \times 18.44 = 37.88 \text{ m}\)

(b) Using model B, the sequence of heights is geometric with the first term 0.96 m and common ratio 0.96. The total distance is twice the sum of the first 20 terms plus the initial drop of 1 m:

\(S_{20} = 0.96 \frac{1 - 0.96^{20}}{1 - 0.96} \approx 23.8 \text{ m}\)

\(Total distance = 1 + 2 \times 23.8 = 48.6 \text{ m}\)

(ii) The infinite sum of the geometric series under model B is:

\(S = \frac{0.96}{1 - 0.96} = 24 \text{ m}\)

Thus, the total distance is 1 + 2 \times 24 = 49 \text{ m}, but since the first bounce is not doubled, the total distance cannot exceed 48 m.

(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.

(i) For a geometric progression, the common ratio is given by \(r = \frac{32}{36} = \frac{8}{9}\).

The sum to infinity for a geometric series is \(S_\infty = \frac{a}{1-r}\), where \(a = 36\) and \(r = \frac{8}{9}\).

Thus, \(S_\infty = \frac{36}{1 - \frac{8}{9}} = \frac{36}{\frac{1}{9}} = 36 \times 9 = 324\).

(ii) For an arithmetic progression, the common difference is \(d = 32 - 36 = -4\).

The sum of an arithmetic series is given by \(S_n = \frac{n}{2} (2a + (n-1)d)\).

Given \(S_n = 0\), we have:

\(0 = \frac{n}{2} (72 + (n-1)(-4))\)

\(0 = \frac{n}{2} (72 - 4n + 4)\)

\(0 = \frac{n}{2} (76 - 4n)\)

\(0 = n(38 - 2n)\)

\(0 = 38n - 2n^2\)

\(2n^2 = 38n\)

\(n(2n - 38) = 0\)

\(n = 0\) or \(n = 19\)

Since \(n = 0\) is not a valid solution, \(n = 19\).