(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, \(n = 19\).

(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: \(0.05 \times 442000 = 22100\).

(i) For an arithmetic progression, the second term is given by:

\(a + d = 96\)

and the fourth term is:

\(a + 3d = 54\)

Subtract the first equation from the second:

\((a + 3d) - (a + d) = 54 - 96\)

\(2d = -42\)

\(d = -21\)

Substitute \(d = -21\) back into \(a + d = 96\):

\(a - 21 = 96\)

\(a = 117\)

(ii) For a geometric progression, the second term is:

\(ar = 96\)

and the fourth term is:

\(ar^3 = 54\)

Divide the second equation by the first:

\(\frac{ar^3}{ar} = \frac{54}{96}\)

\(r^2 = \frac{9}{16}\)

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

Substitute \(r = \frac{3}{4}\) back into \(ar = 96\):

\(a \times \frac{3}{4} = 96\)

\(a = 128\)

Let the first term of the geometric progression be 216 and the common ratio be \(r\).

The fourth term is given by \(216r^3 = 64\).

Solving for \(r\), we have:

\(r^3 = \frac{64}{216} = \frac{8}{27}\)

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

The second term of the geometric progression is \(216r = 144\).

The third term of the geometric progression is \(216r^2 = 96\).

Let the first term of the arithmetic progression be \(a\) and the common difference be \(d\).

The second term of the arithmetic progression is \(a + d = 144\).

The fifth term of the arithmetic progression is \(a + 4d = 96\).

Solving the equations simultaneously:

\(a + d = 144\)

\(a + 4d = 96\)

Subtract the first equation from the second:

\(3d = -48\)

\(d = -16\)

Substitute \(d = -16\) into \(a + d = 144\):

\(a - 16 = 144\)

\(a = 160\)

The sum of the first 21 terms of the arithmetic progression is given by:

\(S_{21} = \frac{21}{2} (2a + 20d)\)

\(S_{21} = \frac{21}{2} (320 + 20(-16))\)

\(S_{21} = \frac{21}{2} (320 - 320)\)

\(S_{21} = 0\)

(i) The first term of the arithmetic progression is 8, so the first term of the geometric progression is also 8. The fifth term of the arithmetic progression is \(8 + 4d\), which is the second term of the geometric progression, so \(8 + 4d = 8r\). The eighth term of the arithmetic progression is \(8 + 7d\), which is the third term of the geometric progression, so \(8 + 7d = 8r^2\).

We have the equations:

\(8 + 4d = 8r\)

\(8 + 7d = 8r^2\)

Eliminating one of the variables, we solve:

\(4r^2 - 7r + 3 = 0\)

Solving this quadratic equation gives \(r = \frac{3}{4}\) and \(d = -\frac{1}{2}\).

(ii) The sum to infinity of the geometric progression is given by \(S_\infty = \frac{a}{1-r}\), where \(a = 8\) and \(r = \frac{3}{4}\).

\(S_\infty = \frac{8}{1 - \frac{3}{4}} = 32\)

(iii) The sum of the first 8 terms of the arithmetic progression is given by \(S_8 = \frac{n}{2} (2a + (n-1)d)\), where \(n = 8\), \(a = 8\), and \(d = -\frac{1}{2}\).

\(S_8 = 4(16 + 7(-\frac{1}{2})) = 50\)

(i) Let the first term be \(a = 81\) and the common ratio be \(r\). The fourth term is given by \(ar^3 = 24\).

Thus, \(81r^3 = 24\) which gives \(r^3 = \frac{24}{81}\).

Solving for \(r\), we get \(r = \frac{2}{3}\) or 0.667.

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

Substituting \(a = 81\) and \(r = \frac{2}{3}\), we get \(S_\infty = \frac{81}{1 - \frac{2}{3}} = 243\).

(iii) The second term of the GP is \(ar = 81 \times \frac{2}{3} = 54\).

The third term of the GP is \(ar^2 = 36\).

These are the first and fourth terms of the AP, so \(3d = -18\) giving \(d = -6\).

The sum of the first ten terms of the AP is \(S_{10} = 5 \times (108 - 54) = 270\).

(i) The general term of an arithmetic progression is given by \(a_n = a + (n-1)d\).

For the 5th term, \(n = 5\):

\(a_5 = a + 4d\).

For the 15th term, \(n = 15\):

\(a_{15} = a + 14d\).

(ii) The terms \(a\), \(a + 4d\), and \(a + 14d\) form a geometric progression.

Thus, \(\frac{a + 4d}{a} = \frac{a + 14d}{a + 4d}\).

Cross-multiplying gives:

\((a + 4d)^2 = a(a + 14d)\).

Expanding both sides:

\(a^2 + 8ad + 16d^2 = a^2 + 14ad\).

Simplifying gives:

\(8ad + 16d^2 = 14ad\).

\(16d^2 = 6ad\).

\(8d = 3a\).

Thus, \(3a = 8d\).

(iii) The common ratio \(r\) of the geometric progression is:

\(r = \frac{a + 4d}{a}\) or \(r = \frac{a + 14d}{a + 4d}\).

Using \(r = \frac{a + 4d}{a}\):

\(r = \frac{a + 4d}{a} = 1 + \frac{4d}{a}\).

Substituting \(3a = 8d\), we have \(a = \frac{8d}{3}\).

\(r = 1 + \frac{4d}{\frac{8d}{3}} = 1 + \frac{12d}{8d} = 1 + \frac{3}{2} = 2.5\).

(i) Let the first term of the geometric progression be \(a\) and the common ratio be \(r\). The second term is given by \(ar = 3\). The sum to infinity of a geometric progression is given by \(\frac{a}{1-r} = 12\).

We have the equations:

\(ar = 3\)

\(\frac{a}{1-r} = 12\)

From \(ar = 3\), we can express \(r\) as \(r = \frac{3}{a}\).

Substitute \(r = \frac{3}{a}\) into \(\frac{a}{1-r} = 12\):

\(\frac{a}{1-\frac{3}{a}} = 12\)

\(\frac{a^2}{a-3} = 12\)

\(a^2 = 12(a-3)\)

\(a^2 = 12a - 36\)

\(a^2 - 12a + 36 = 0\)

Solving this quadratic equation, we find \(a = 6\).

(ii) For the arithmetic progression, the first term \(a = 6\) and the second term is 3, so the common difference \(d = 3 - 6 = -3\).

The sum of the first 20 terms of an arithmetic progression is given by:

\(S_{20} = \frac{n}{2} (2a + (n-1)d)\)

\(S_{20} = \frac{20}{2} (2 \times 6 + 19 \times (-3))\)

\(S_{20} = 10 (12 - 57)\)

\(S_{20} = 10 \times (-45)\)

\(S_{20} = -450\)

(i) The profit follows a geometric progression with initial term \(a = 250,000\) and common ratio \(r = 1.05\). The year 2008 is the 9th term, so we calculate \(ar^8 = 250,000 \times 1.05^8 = 369,000\).

(ii) The total profit for 10 years is given by the sum of a geometric series: \(S_{10} = 250,000 \left( \frac{1.05^{10} - 1}{0.05} \right) = 3,140,000\).

(iii) Under plan B, the profit forms an arithmetic progression. The sum of the profits for 10 years is \(S_{10} = 5(500,000 + 9D)\). Setting this equal to the total from plan A, \(5(500,000 + 9D) = 3,140,000\), solving gives \(D = 14,300\).

For the geometric progression (GP), the first term is given as \(a = 192\), the common ratio \(r = 1.5\), and the number of terms \(n = 6\).

The sum of the GP is given by:

\(S_6 = a \frac{r^n - 1}{r - 1} = 192 \frac{1.5^6 - 1}{1.5 - 1}\)

Calculating this gives:

\(S_6 = 192 \times 0.5 \times (1.5^6 - 1) = 3990\)

For the arithmetic progression (AP), the number of terms \(n = 21\), and the common difference \(d = 1.5\).

The sum of the AP is given by:

\(S_{21} = \frac{n}{2} (2a + (n-1)d) = \frac{21}{2} (2a + 20 \times 1.5)\)

Equating the sums of the GP and AP:

\(3990 = \frac{21}{2} (2a + 30)\)

Solving for \(a\):

\(3990 = 21(a + 15)\)

\(3990 = 21a + 315\)

\(21a = 3675\)

\(a = 175\)

The first term of the AP is 175.

The last term of the AP is given by:

\(l = a + 20d = 175 + 20 \times 1.5 = 205\)

Thus, the first term is 175 and the last term is 205.

(i) For the arithmetic progression, the first term is \(a = 12\) and the fifth term is \(a + 4d = 18\). Solving for \(d\), we have:

\(a + 4d = 18\)

\(12 + 4d = 18\)

\(4d = 6\)

\(d = 1.5\)

The sum of the first 25 terms \(S_{25}\) is given by:

\(S_{25} = \frac{25}{2} (2a + (25-1)d)\)

\(S_{25} = \frac{25}{2} (24 + 24 \times 1.5)\)

\(S_{25} = 750\)

(ii) For the geometric progression, the first term is \(a = 12\) and the fifth term is \(ar^4 = 18\). Solving for \(r\), we have:

\(ar^4 = 18\)

\(12r^4 = 18\)

\(r^4 = 1.5\)

The 13th term \(ar^{12}\) is given by:

\(ar^{12} = 12 \times (1.5)^3\)

\(ar^{12} = 40.5 \text{ or } 40.6\)

Let the common ratio of the geometric progression be \(r\) and the common difference of the arithmetic progression be \(d\).

The terms of the geometric progression are \(a, ar, ar^2, ar^3, ar^4, \ldots\).

The terms of the arithmetic progression are \(a, a+d, a+2d, a+3d, a+4d, \ldots\).

From the problem, we have:

1. \(ar^2 = a + d\)

2. \(ar^4 = a + 5d\)

From equation 1: \(ar^2 = a + d\)

From equation 2: \(ar^4 = a + 5d\)

We can express \(ar^4\) in terms of \(ar^2\):

\(ar^4 = (ar^2)^2 = (a + d)^2\)

Equating the two expressions for \(ar^4\):

\(a(a + 5d) = (a + d)^2\)

Expanding both sides:

\(a^2 + 5ad = a^2 + 2ad + d^2\)

Simplifying gives:

\(3ad = d^2\)

\(d(3a - d) = 0\)

Since \(d \neq 0\), we have \(d = 3a\).

Substitute \(d = 3a\) into the sum formula for the arithmetic progression:

The sum of the first 20 terms is:

\(S_{20} = \frac{20}{2} [2a + 19 \times 3a]\)

\(S_{20} = 10 [2a + 57a]\)

\(S_{20} = 10 \times 59a\)

\(S_{20} = 590a\)

(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:

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

(i) The growth in the first day is given by:

\(41.2 - 40 = 1.2 \text{ cm}\)

Assuming this daily growth continues for 60 days, the predicted height is:

\(40 + 60 \times 1.2 = 112 \text{ cm}\)

(ii) The daily percentage growth rate is:

\(\frac{41.2}{40} = 1.03\)

Assuming this percentage growth continues, the predicted height after 60 days is:

\(40 \times (1.03)^{60} \approx 236 \text{ cm}\)