\(The first term of the geometric progression is a = p and the common ratio is r = 2.\)

The sum of the first n terms of a geometric progression is given by:

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

Substituting the values, we have:

\(S_n = \frac{p(2^n - 1)}{2 - 1} = p(2^n - 1)\)

We are given that:

\(p(2^n - 1) > 1000p\)

Dividing both sides by p (since p is positive), we get:

\(2^n - 1 > 1000\)

Adding 1 to both sides gives:

\(2^n > 1001\)

The first term of the geometric series is given as 6, and the second term is 2. The common ratio \(r\) can be found by dividing the second term by the first term:

\(r = \frac{2}{6} = \frac{1}{3}\)

The formula for the sum to infinity of a geometric series is:

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

where \(a\) is the first term and \(|r| < 1\).

Substituting the known values:

\(S_\infty = \frac{6}{1 - \frac{1}{3}} = \frac{6}{\frac{2}{3}} = 6 \times \frac{3}{2} = 9\)

Thus, the sum to infinity is 9.

The formula for the sum of the first n terms of a geometric progression is given by:

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

where \(a\) is the first term and \(r\) is the common ratio.

The formula for the sum to infinity of a geometric progression is:

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

Given \(r = 0.99\), we need to find \(\frac{S_{100}}{S_\infty} \times 100\).

First, calculate \(1 - r^n\) for \(n = 100\):

\(1 - 0.99^{100} \approx 0.634\)

Then, the ratio \(\frac{S_{100}}{S_\infty}\) is:

\(\frac{a(1-0.99^{100})}{a} = 1 - 0.99^{100} \approx 0.634\)

Expressing this as a percentage:

\(0.634 \times 100 \approx 63.4\%\)

Rounding to 2 significant figures gives:

63%

(i) The amount of salt obtained each week forms a geometric progression with the first term \(a = 8000\) kg and a common ratio \(r = 1.02\). The amount of salt obtained in the 12th week is given by the formula for the \(n\)-th term of a geometric progression:

\(a_n = a imes r^{n-1}\)

Substituting the values, we have:

\(a_{12} = 8000 imes (1.02)^{11}\)

\(a_{12} \approx 9950 \text{ kg}\)

(ii) The total amount of salt obtained in the first 12 weeks is the sum of the first 12 terms of the geometric progression. The sum \(S_n\) of the first \(n\) terms of a geometric progression is given by:

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

Substituting the values, we have:

\(S_{12} = 8000 \frac{(1.02)^{12} - 1}{1.02 - 1}\)

\(S_{12} \approx 107000 \text{ kg}\)

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

The second term is given by \(ar = 12\).

The sum to infinity is given by \(\frac{a}{1-r} = 54\).

From \(ar = 12\), we have \(a = \frac{12}{r}\).

Substitute \(a = \frac{12}{r}\) into \(\frac{a}{1-r} = 54\):

\(\frac{\frac{12}{r}}{1-r} = 54\)

\(\frac{12}{r(1-r)} = 54\)

\(12 = 54r(1-r)\)

\(12 = 54r - 54r^2\)

\(54r^2 - 54r + 12 = 0\)

Divide through by 6:

\(9r^2 - 9r + 2 = 0\)

Using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 9\), \(b = -9\), \(c = 2\):

\(r = \frac{9 \pm \sqrt{(-9)^2 - 4 \times 9 \times 2}}{2 \times 9}\)

\(r = \frac{9 \pm \sqrt{81 - 72}}{18}\)

\(r = \frac{9 \pm \sqrt{9}}{18}\)

\(r = \frac{9 \pm 3}{18}\)

\(r = \frac{12}{18} = \frac{2}{3}\) or \(r = \frac{6}{18} = \frac{1}{3}\)

For \(r = \frac{2}{3}\), \(a = \frac{12}{\frac{2}{3}} = 18\).

For \(r = \frac{1}{3}\), \(a = \frac{12}{\frac{1}{3}} = 36\).

Thus, the possible values of the first term are 18 or 36.