(i) For the geometric progression (GP), the terms are \(8, 8r, 8r^2\).

For the arithmetic progression (AP), the terms are \(8, 8 + 8d, 8 + 20d\).

Equating the terms: \(8r = 8 + 8d\) and \(8r^2 = 8 + 20d\).

From \(8r = 8 + 8d\), we get \(r = 1 + d\).

Substitute \(d = r - 1\) into \(8r^2 = 8 + 20d\):

\(8r^2 = 8 + 20(r - 1)\)

\(8r^2 = 8 + 20r - 20\)

\(8r^2 = 20r - 12\)

\(8r^2 - 20r + 12 = 0\)

Divide by 4: \(2r^2 - 5r + 3 = 0\)

Solving the quadratic: \((2r - 3)(r - 1) = 0\)

\(r = 1.5\) (since \(r \neq 1\))

(ii) The 4th term of the GP is \(ar^3 = 8 \times (1.5)^3 = 27\).

For the AP, \(d = 0.5\) (since \(r = 1.5\) and \(r = 1 + d\)).

The 4th term of the AP is \(a + 3d = 8 + 3 \times 0.5 = 9.5\).

Let the first term of the geometric progression (GP) be \(a = 64\) and the common ratio be \(r\). The second term is \(ar = 48\), so \(64r = 48\), giving \(r = \frac{3}{4}\).

The third term of the GP is \(ar^2 = 64 \times \left(\frac{3}{4}\right)^2 = 36\).

For the arithmetic progression (AP), the first term \(a = 64\) and the ninth term \(a + 8d = 48\). Solving for \(d\), we have:

\(64 + 8d = 48\)

\(8d = 48 - 64 = -16\)

\(d = -2\)

The nth term of the AP is \(a + (n-1)d = 36\). Substituting the known values:

\(64 + (n-1)(-2) = 36\)

\(64 - 2(n-1) = 36\)

\(64 - 2n + 2 = 36\)

\(66 - 2n = 36\)

\(2n = 66 - 36 = 30\)

\(n = 15\)

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

For the first 9 terms, \(S_9 = \frac{9}{2} (24 + 8d) = 135\).

Solving for \(d\), we have:

\(9(24 + 8d) = 270\)

\(24 + 8d = 30\)

\(8d = 6\)

\(d = \frac{3}{4}\).

(ii) The 9th term of the arithmetic progression is \(12 + 8 \times \frac{3}{4} = 18\).

In the geometric progression, the first term is 12, the second term is 18.

The common ratio \(r\) is \(\frac{18}{12} = \frac{3}{2}\).

The third term of the geometric progression is \(ar^2 = 12 \times \left(\frac{3}{2}\right)^2 = 27\).

The \(n\)th term of the arithmetic progression is \(12 + (n-1) \times \frac{3}{4} = 27\).

Solving for \(n\), we have:

\(12 + \frac{3}{4}(n-1) = 27\)

\(\frac{3}{4}(n-1) = 15\)

\(n-1 = 20\)

\(n = 21\).

(i) For an arithmetic progression, the first term is \(a = 4\) and the common difference \(d = 8 - 4 = 4\). The sum of the first \(n\) terms is given by:

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

Substituting \(n = 10\), \(a = 4\), and \(d = 4\):

\(S_{10} = \frac{10}{2} (2 \times 4 + 9 \times 4) = 5 (8 + 36) = 5 \times 44 = 220\)

(ii) For a geometric progression, the first term is \(a = 4\) and the common ratio \(r = \frac{8}{4} = 2\). The sum of the first \(n\) terms is given by:

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

Substituting \(n = 10\), \(a = 4\), and \(r = 2\):

\(S_{10} = 4 \frac{2^{10} - 1}{2 - 1} = 4 (1024 - 1) = 4 \times 1023 = 4092\)

(i) Using Model 1, the prize money increases by $1000 each day, forming an arithmetic series: 1000, 2000, 3000, ..., up to 40 terms.

The sum of an arithmetic series is given by \(S_n = \frac{n}{2} (a + l)\), where \(n\) is the number of terms, \(a\) is the first term, and \(l\) is the last term.

Here, \(n = 40\), \(a = 1000\), and \(l = 40000\).

\(S_{40} = \frac{40}{2} (1000 + 40000) = 20 \times 41000 = 820000\).

The total donation is 5% of this sum: \(0.05 \times 820000 = 41000\).

(ii) Using Model 2, the prize money increases by 10% each day, forming a geometric series: 1000, 1000 \(\times 1.1\), 1000 \(\times 1.1^2\), ..., up to 40 terms.

The sum of a geometric series is given by \(S_n = a \frac{r^n - 1}{r - 1}\), where \(a\) is the first term and \(r\) is the common ratio.

Here, \(a = 1000\), \(r = 1.1\), and \(n = 40\).

\(S_{40} = 1000 \frac{1.1^{40} - 1}{1.1 - 1} \approx 442000\).

The total donation is 5% of this sum: \(0.05 \times 442000 = 22100\).