(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:

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

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

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

\(S_{20} = 590a\)