Given the terms of the geometric progression: \(a = 2p + 6\), \(ar = -2p\), and \(ar^2 = p + 2\).

Using the relationship \(b^2 = ac\), we have:

\((-2p)^2 = (2p + 6) \times (p + 2)\)

\(4p^2 = (2p + 6)(p + 2)\)

Expanding and simplifying:

\(4p^2 = 2p^2 + 4p + 12p + 12\)

\(4p^2 = 2p^2 + 16p + 12\)

\(2p^2 - 16p - 12 = 0\)

Solving this quadratic equation for \(p\):

\(p^2 - 8p - 6 = 0\)

Using the quadratic formula \(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(p = \frac{8 \pm \sqrt{64 + 24}}{2}\)

\(p = \frac{8 \pm \sqrt{88}}{2}\)

\(p = \frac{8 \pm 2\sqrt{22}}{2}\)

\(p = 4 \pm \sqrt{22}\)

Since \(p\) is positive, we take \(p = 6\).

Substituting \(p = 6\) back into the expressions for \(a\) and \(r\):

\(a = 2(6) + 6 = 18\)

\(r = \frac{-2(6)}{18} = -\frac{2}{3}\)

The sum to infinity of a geometric progression is given by:

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

\(S_\infty = \frac{18}{1 - (-\frac{2}{3})}\)

\(S_\infty = \frac{18}{1 + \frac{2}{3}}\)

\(S_\infty = \frac{18}{\frac{5}{3}}\)

\(S_\infty = 18 \times \frac{3}{5}\)

\(S_\infty = 10.8\)

(a) The sum to infinity of a geometric progression is given by \(S = \frac{a}{1-r}\) and \(2S = \frac{a}{1-R}\).

Equating the sums: \(\frac{a}{1-r} = \frac{2a}{1-R}\).

Cross-multiplying gives: \(a(1-R) = 2a(1-r)\).

Expanding and simplifying: \(a - aR = 2a - 2ar\).

Rearranging gives: \(2ar - aR = a\).

Dividing by \(a\) (assuming \(a \neq 0\)): \(2r - R = 1\).

Thus, \(r = 2R - 1\).

(b) The 3rd term of the first progression is \(ar^2\) and the 2nd term of the second progression is \(aR\).

Given \(ar^2 = aR\), we have \(r^2 = R\).

Substituting \(r = 2R - 1\) into \(r^2 = R\):

\((2R - 1)^2 = R\).

Expanding gives: \(4R^2 - 4R + 1 = R\).

Rearranging: \(4R^2 - 5R + 1 = 0\).

Solving this quadratic equation using the quadratic formula:

\(R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 4, b = -5, c = 1\).

\(R = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8}\).

Thus, \(R = 1\) or \(R = \frac{1}{4}\).

For \(R = \frac{1}{4}\), substituting back gives \(S = \frac{a}{1-r} = \frac{a}{1-(2 \times \frac{1}{4} - 1)} = \frac{a}{\frac{1}{2}} = 2a\).

Since \(2S = 2a\), \(S = \frac{2a}{3}\).

(a) The selling price of the necklace increases by 5% each year. This can be modeled by the formula for compound interest:

\(P_n = 36000 \times (1.05)^n\)

To find the selling price in 2008, we calculate for \(n = 8\) (since 2008 is 8 years after 2000):

\(P_8 = 36000 \times (1.05)^8\)

Calculating this gives:

\(P_8 = 36000 \times 1.477455 \approx 53200\)

So, the selling price in 2008 is $53,200.

(b) The total amount of money obtained over 10 years is the sum of a geometric series where the first term \(a = 36000\) and the common ratio \(r = 1.05\).

The sum of the first \(n\) terms of a geometric series is given by:

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

For \(n = 10\):

\(S_{10} = 36000 \frac{(1.05)^{10} - 1}{1.05 - 1}\)

Calculating this gives:

\(S_{10} = 36000 \frac{1.628895 - 1}{0.05}\)

\(S_{10} = 36000 \times 12.5779 \approx 453000\)

So, the total amount of money obtained in the ten-year period is $453,000.

(a) The bonus for the first year is $600. The basic salary is $30,000. To find the bonus as a percentage of the basic salary:

\(\text{Percentage} = \left( \frac{600}{30000} \right) \times 100 = 2\%\)

(b) The bonus at the end of the 24th year is calculated as follows:

\(Initial bonus = $600\)

\(Increase over 23 years = 23 \times $100 = $2300\)

\(Total bonus = $600 + $2300 = $2900\)

The basic salary at the end of the 24th year is:

\(\text{Salary} = 30000 \times 1.03^{23} \approx 59207.60\)

Express the bonus as a percentage of the salary:

\(\text{Percentage} = \left( \frac{2900}{59207.60} \right) \times 100 \approx 4.9\%\)

(i) For a geometric progression, the ratio between consecutive terms is constant. Therefore, \(\frac{5k - 6}{3k} = \frac{6k - 4}{5k - 6}\).

Cross-multiplying gives \((5k - 6)^2 = 3k(6k - 4)\).

Expanding both sides: \(25k^2 - 60k + 36 = 18k^2 - 12k\).

Rearranging gives \(7k^2 - 48k + 36 = 0\).

(ii) Solving \(7k^2 - 48k + 36 = 0\) gives \(k = \frac{6}{7}\) or \(k = 6\).

When \(k = \frac{6}{7}\), the first term is \(3 \times \frac{6}{7} = \frac{18}{7}\) and the common ratio \(r = \frac{2}{3}\).

When \(k = 6\), the common ratio \(r = \frac{4}{3}\).

(iii) The progression is convergent if \(|r| < 1\). Thus, \(r = \frac{2}{3}\) is convergent.

The sum to infinity is \(S_\infty = \frac{a}{1-r}\) where \(a = \frac{18}{7}\) and \(r = \frac{2}{3}\).

\(S_\infty = \frac{\frac{18}{7}}{1 - \frac{2}{3}} = \frac{18}{7} \times 3 = \frac{54}{35}\) or 1.54.