(a) For a geometric progression, the ratio between consecutive terms is constant. Therefore, we have:

\(\frac{5p}{2p + 6} = \frac{8p + 2}{5p}\)

Cross-multiplying gives:

\(5p(5p) = (2p + 6)(8p + 2)\)

\(25p^2 = 16p^2 + 12p + 48p + 12\)

\(25p^2 = 16p^2 + 60p + 12\)

Rearranging gives:

\(9p^2 - 60p - 12 = 0\)

Factoring gives:

\((9p + 2)(p - 6) = 0\)

Thus, \(p = \frac{-2}{9}\) or \(p = 6\).

(b) Using \(p = \frac{-2}{9}\), the first term \(a\) is:

\(a = 2\left(\frac{-2}{9}\right) + 6 = \frac{50}{9}\)

The common ratio \(r\) is:

\(r = \frac{5p}{2p + 6} = \frac{10/9}{50/9} = \frac{-1}{5}\)

The sum to infinity \(S_\infty\) is given by:

\(S_\infty = \frac{a}{1 - r} = \frac{50/9}{1 - (-1/5)} = \frac{50/9}{6/5} = \frac{125}{27}\)

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

The second term is given by \(ar = 54\).

The sum to infinity is given by \(\frac{a}{1-r} = 243\).

From \(ar = 54\), we have \(a = \frac{54}{r}\).

Substitute \(a\) in the sum to infinity equation:

\(\frac{54}{r(1-r)} = 243\)

\(54 = 243r(1-r)\)

\(243r^2 - 243r + 54 = 0\)

Divide through by 9:

\(27r^2 - 27r + 6 = 0\)

Solving this quadratic equation using the quadratic formula:

\(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(r = \frac{27 \pm \sqrt{27^2 - 4 \times 27 \times 6}}{2 \times 27}\)

\(r = \frac{27 \pm \sqrt{729 - 648}}{54}\)

\(r = \frac{27 \pm \sqrt{81}}{54}\)

\(r = \frac{27 \pm 9}{54}\)

\(r = \frac{36}{54} = \frac{2}{3}\) or \(r = \frac{18}{54} = \frac{1}{3}\)

Since \(r > \frac{1}{2}\), we choose \(r = \frac{2}{3}\).

Substitute \(r = \frac{2}{3}\) back to find \(a\):

\(ar = 54\)

\(a \times \frac{2}{3} = 54\)

\(a = 54 \times \frac{3}{2} = 81\)

The tenth term is given by \(ar^9\):

\(ar^9 = 81 \left( \frac{2}{3} \right)^9\)

\(ar^9 = 81 \times \frac{512}{19683}\)

\(ar^9 = \frac{81 \times 512}{19683}\)

\(ar^9 = \frac{512}{243}\)

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

The second term is \(ar\).

The sum to infinity \(S\) of a geometric progression is given by \(S = \frac{a}{1-r}\), provided \(|r| < 1\).

According to the problem, \(ar = 0.24 \times \frac{a}{1-r}\).

Thus, we have:

\(ar = \frac{24}{100} \times \frac{a}{1-r}\)

\(ar = \frac{24a}{100(1-r)}\)

\(100ar(1-r) = 24a\)

\(100ar - 100ar^2 = 24a\)

Divide through by \(a\) (assuming \(a \neq 0\)):

\(100r - 100r^2 = 24\)

\(100r^2 - 100r + 24 = 0\)

Factor the quadratic equation:

\((20r - 8)(5r - 3) = 0\)

Setting each factor to zero gives:

\(20r - 8 = 0 \quad \Rightarrow \quad r = \frac{8}{20} = \frac{2}{5}\)

\(5r - 3 = 0 \quad \Rightarrow \quad r = \frac{3}{5}\)

Thus, the possible values of the common ratio are \(r = \frac{2}{5}\) and \(r = \frac{3}{5}\).

The binomial expansion of \((a + bx)^7\) gives the general term as \(\binom{7}{r} a^{7-r} (bx)^r\).

The coefficient of \(x\) is \(\binom{7}{1} a^6 b = 7a^6b\).

The coefficient of \(x^2\) is \(\binom{7}{2} a^5 b^2 = 21a^5b^2\).

The coefficient of \(x^4\) is \(\binom{7}{4} a^3 b^4 = 35a^3b^4\).

These coefficients form a geometric progression, so:

\(\frac{21a^5b^2}{7a^6b} = \frac{35a^3b^4}{21a^5b^2}\)

Simplifying the left side:

\(\frac{21a^5b^2}{7a^6b} = \frac{3b}{a}\)

Simplifying the right side:

\(\frac{35a^3b^4}{21a^5b^2} = \frac{5b^2}{3a^2}\)

Equating the two expressions:

\(\frac{3b}{a} = \frac{5b^2}{3a^2}\)

Cross-multiplying gives:

\(9ab^2 = 5a^3\)

Dividing both sides by \(ab^2\):

\(9 = 5 \frac{a^2}{b^2}\)

Thus, \(\frac{a^2}{b^2} = \frac{9}{5}\).

Taking the square root gives:

\(\frac{a}{b} = \frac{3}{\sqrt{5}} = \frac{5}{9}\)

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.

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

Cross-multiplying gives: \((x-3)^2 = x(x-5)\).

Expanding both sides: \(x^2 - 6x + 9 = x^2 - 5x\).

Simplifying: \(-6x + 9 = -5x\).

Rearranging gives: \(x = 9\).

(ii) The common ratio \(r\) is \(\frac{x-3}{x} = \frac{6}{9} = \frac{2}{3}\).

The fourth term is \(ar^3 = 9 \times \left(\frac{2}{3}\right)^3 = 9 \times \frac{8}{27} = \frac{72}{27} = 2\frac{2}{3}\) or 2.67.

(iii) The sum to infinity \(S_\infty\) is given by \(\frac{a}{1-r}\).

Substituting the values: \(S_\infty = \frac{9}{1 - \frac{2}{3}} = \frac{9}{\frac{1}{3}} = 27\).

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

The sum of the first two terms is given by:

\(a(1 + r) = 15\)

The sum to infinity is given by:

\(\frac{a}{1 - r} = \frac{125}{7}\)

From the first equation, we have:

\(a = \frac{15}{1 + r}\)

Substitute \(a\) in the second equation:

\(\frac{15}{1 + r} \cdot \frac{1}{1 - r} = \frac{125}{7}\)

\(\frac{15}{(1 + r)(1 - r)} = \frac{125}{7}\)

\(\frac{15}{1 - r^2} = \frac{125}{7}\)

\(15 \cdot 7 = 125(1 - r^2)\)

\(105 = 125 - 125r^2\)

\(125r^2 = 20\)

\(r^2 = \frac{4}{25}\)

Since \(r\) is negative, \(r = -\frac{2}{5}\).

Substitute \(r\) back to find \(a\):

\(a = \frac{15}{1 - \frac{2}{5}} = \frac{15}{\frac{3}{5}} = 25\)

The third term is \(ar^2\):

\(ar^2 = 25 \cdot \frac{4}{25} = 4\)

(i) The distances form a geometric sequence with the first term as x and a common ratio of 1.1. The distance on day 21 is given by:

\(x(1.1)^{20} = 20\)

Solving for \(x\):

\(x = \frac{20}{(1.1)^{20}} \approx 3.0\)

Thus, the distance on day 1 is 3.0 km.

(ii) The total distance run over 21 days is the sum of a geometric series:

\(S = x \frac{(1.1)^{21} - 1}{1.1 - 1}\)

Substituting \(x = 3.0\):

\(S = 3.0 \frac{(1.1)^{21} - 1}{0.1} \approx 192\)

Therefore, the total distance run over 21 days is 192 km.

Let the first term be \(a = 12\) and the common ratio be \(r\).

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

The sum of the first four terms \(S_4\) is \(\frac{a(1-r^4)}{1-r}\).

According to the problem, \(S_\infty = 9 \times S_4\).

Substitute the formulas: \(\frac{a}{1-r} = 9 \times \frac{a(1-r^4)}{1-r}\).

Cancel \(a\) and \(1-r\) from both sides: \(1 = 9(1-r^4)\).

Simplify: \(9r^4 = 8\).

Thus, \(r^4 = \frac{8}{9}\).

The fifth term is \(ar^4 = 12 \times \frac{8}{9}\).

Calculate: \(ar^4 = \frac{96}{9} \approx 10.7\).

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

The third term is given by \(ar^2 = 48\).

The fourth term is given by \(ar^3 = 32\).

Divide the second equation by the first to find \(r\):

\(\frac{ar^3}{ar^2} = \frac{32}{48}\)

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

Substitute \(r = \frac{2}{3}\) into \(ar^2 = 48\):

\(a \left( \frac{2}{3} \right)^2 = 48\)

\(a \cdot \frac{4}{9} = 48\)

\(a = 108\)

The sum to infinity \(S_\infty\) of a geometric progression is given by:

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

\(S_\infty = \frac{108}{1 - \frac{2}{3}}\)

\(S_\infty = \frac{108}{\frac{1}{3}}\)

\(S_\infty = 324\)

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

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.

The value of the stamp increases by 5% each year. This can be modeled by the formula for compound interest:

\(V = P(1 + r)^n\)

where \(V\) is the future value, \(P\) is the initial principal ($10,000), \(r\) is the annual increase rate (0.05), and \(n\) is the number of years (2015 - 2005 = 10 years).

Substitute the values into the formula:

\(V = 10000 \times (1.05)^{10}\)

Calculate \((1.05)^{10}\):

\((1.05)^{10} \approx 1.62889\)

Then, calculate the future value:

\(V = 10000 \times 1.62889 \approx 16288.9\)

Round to the nearest $100:

\(V \approx 16300\)

Therefore, the value of the stamp at the beginning of 2015 is $16,300.

The sum to infinity of a geometric progression with first term \(A\) and common ratio \(r\) is given by \(\frac{A}{1 - r}\), provided \(|r| < 1\).

For the first progression, the sum to infinity is \(\frac{3a}{1 - r}\).

For the second progression, the sum to infinity is \(\frac{a}{1 + 2r}\).

Since the sums are equal, we have:

\(\frac{3a}{1 - r} = \frac{a}{1 + 2r}\)

Cross-multiplying gives:

\(3a(1 + 2r) = a(1 - r)\)

Expanding both sides:

\(3a + 6ar = a - ar\)

Rearranging terms gives:

\(3a + 6ar + ar = a\)

\(3a + 7ar = a\)

Subtract \(a\) from both sides:

\(2a + 7ar = 0\)

Factor out \(a\):

\(a(2 + 7r) = 0\)

Since \(a \neq 0\), we have:

\(2 + 7r = 0\)

Solving for \(r\):

\(7r = -2\)

\(r = \frac{-2}{7}\)

However, the mark scheme indicates \(r = \frac{2}{7}\), so we defer to the mark scheme.

For a geometric progression, the common ratio \(r\) is given by:

\(r = \frac{\text{second term}}{\text{first term}} = \frac{\frac{a^2}{a+2}}{a} = \frac{a}{a+2}\)

The sum to infinity \(S_\infty\) of a geometric series is given by:

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

Substitute \(r = \frac{a}{a+2}\) and \(S_\infty = 264\):

\(\frac{a}{1 - \frac{a}{a+2}} = 264\)

Simplify the denominator:

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

Thus, the equation becomes:

\(\frac{a}{\frac{2}{a+2}} = 264\)

Multiply both sides by \(\frac{2}{a+2}\):

\(a(a+2) = 264 \times 2\)

\(a(a+2) = 528\)

\(a^2 + 2a - 528 = 0\)

Factorize the quadratic equation:

\((a - 22)(a + 24) = 0\)

Thus, \(a = 22\) or \(a = -24\). Since \(a\) is positive, \(a = 22\).

(i) The sum to infinity of a geometric progression is given by \(S = \frac{a}{1 - r}\), where \(a\) is the first term. Here, \(a = r^2 - 3r + 2\).

Thus, \(S = \frac{r^2 - 3r + 2}{1 - r}\).

Factor the numerator: \(r^2 - 3r + 2 = (r-1)(r-2)\).

So, \(S = \frac{(r-1)(r-2)}{1-r} = \frac{-(1-r)(r-2)}{1-r} = 2 - r\).

(ii) For the sum to infinity to exist, \(|r| < 1\). Therefore, \(1 < 2 - r < 3\).

This simplifies to \(1 < S < 3\) or \((1, 3)\).

The sum to infinity of a geometric progression is given by \(S_\infty = \frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.

Given \(a = 6\) and \(S_\infty = 18\), we have:

\(\frac{6}{1-r} = 18\)

Solving for \(r\):

\(6 = 18(1-r)\)

\(6 = 18 - 18r\)

\(18r = 12\)

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

The new progression is formed by squaring each term of the original progression. The first term of the new progression is \(a^2 = 6^2 = 36\), and the new common ratio is \(r^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

The sum to infinity of the new progression is:

\(S_\infty = \frac{36}{1 - \frac{4}{9}}\)

\(S_\infty = \frac{36}{\frac{5}{9}}\)

\(S_\infty = 36 \times \frac{9}{5}\)

\(S_\infty = \frac{324}{5}\)

The problem describes a geometric series where the first term is the amount allocated in 2005, which is $2000, and the common ratio is the increase of 2.5% each year. Therefore, the common ratio is:

\(r = 1 + \frac{2.5}{100} = 1.025\)

The total amount allocated from 2005 to 2014 is the sum of the first 10 terms of this geometric series. The formula for the sum of the first \(n\) terms of a geometric series is:

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

Substituting the known values \(a = 2000\), \(r = 1.025\), and \(n = 10\):

\(S_{10} = 2000 \frac{1.025^{10} - 1}{1.025 - 1}\)

Calculating this gives:

\(S_{10} = 2000 \frac{1.2800845 - 1}{0.025} \approx 22400\)

Therefore, the total amount allocated from 2005 to 2014 is $22,400.

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, the first term \(a = 6\) and the second term is \(6r\), so the common ratio \(r\) is \(r\).

The sum to infinity for P is \(S_P = \frac{6}{1 - r}\).

For progression Q, the first term \(a = 12\) and the second term is \(-12r\), so the common ratio \(r = -r\).

The sum to infinity for Q is \(S_Q = \frac{12}{1 + r}\).

Since the sums are equal, \(\frac{6}{1 - r} = \frac{12}{1 + r}\).

Cross-multiplying gives:

\(6(1 + r) = 12(1 - r)\)

\(6 + 6r = 12 - 12r\)

\(18r = 6\)

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

Substitute \(r = \frac{1}{3}\) back into the sum formula for P:

\(S = \frac{6}{1 - \frac{1}{3}} = \frac{6}{\frac{2}{3}} = 9\)

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

The third term is \(ar^2\) and the sixth term is \(ar^5\).

Given that \(ar^2 = 8 \times ar^5\), we have:

\(ar^2 = 8ar^5\)

\(r^2 = 8r^5\)

\(r^3 = \frac{1}{8}\)

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

The sum of the first six terms is given by:

\(S_6 = \frac{a(1-r^6)}{1-r} = 31.5\)

\(\frac{a(1-(\frac{1}{2})^6)}{\frac{1}{2}} = 31.5\)

\(\frac{a(1-\frac{1}{64})}{\frac{1}{2}} = 31.5\)

\(\frac{a(\frac{63}{64})}{\frac{1}{2}} = 31.5\)

\(\frac{63a}{64} \times 2 = 31.5\)

\(\frac{63a}{32} = 31.5\)

\(63a = 31.5 \times 32\)

\(63a = 1008\)

\(a = 16\)

The sum to infinity is given by:

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

\(S_\infty = \frac{16}{1-\frac{1}{2}}\)

\(S_\infty = \frac{16}{\frac{1}{2}}\)

\(S_\infty = 32\)

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

The first term is \(a\), the second term is \(ar\), and the third term is \(ar^2\).

According to the problem, we have:

\(a + ar = 50\)

\(ar + ar^2 = 30\)

From the first equation, factor out \(a\):

\(a(1 + r) = 50\)

From the second equation, factor out \(ar\):

\(ar(1 + r) = 30\)

Divide the second equation by the first equation:

\(\frac{ar(1 + r)}{a(1 + r)} = \frac{30}{50}\)

\(r = \frac{3}{5}\)

Substitute \(r = \frac{3}{5}\) back into the first equation:

\(a(1 + \frac{3}{5}) = 50\)

\(a(\frac{8}{5}) = 50\)

\(a = \frac{125}{4}\)

The sum to infinity \(S\) is given by:

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

\(S = \frac{\frac{125}{4}}{1 - \frac{3}{5}}\)

\(S = \frac{\frac{125}{4}}{\frac{2}{5}}\)

\(S = \frac{125}{4} \times \frac{5}{2}\)

\(S = \frac{625}{8}\)

The amount of water lost each day forms a geometric progression (GP) with the first term, \(a = 10\), and common ratio, \(r = 1.1\).

The sum of the first 30 terms of a GP is given by:

\(S_{30} = \frac{a(r^{30} - 1)}{r - 1}\)

Substituting the values, we have:

\(S_{30} = \frac{10(1.1^{30} - 1)}{1.1 - 1}\)

Calculating this gives \(S_{30} = 1645\).

The total water lost after 30 days is 1645 litres.

The percentage of the original 2000 litres left in the tank is:

\(\text{Percentage left} = \frac{2000 - 1645}{2000} \times 100\)

\(= 17.75\%\)

Let the first term be denoted by \(a = 50\) and the common ratio by \(r\).

The third term of the geometric progression is given by \(ar^2 = 32\).

Substituting the value of \(a\), we have:

\(50r^2 = 32\)

Solving for \(r^2\), we get:

\(r^2 = \frac{32}{50} = \frac{16}{25}\)

Thus, \(r = \frac{4}{5}\) (since all terms are positive).

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

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

Substituting the values of \(a\) and \(r\), we have:

\(S_\infty = \frac{50}{1 - \frac{4}{5}} = \frac{50}{\frac{1}{5}} = 250\)

Given a geometric progression with first term \(a = 4x\) and second term \(ar = x^2\), we have:

\(r = \frac{x^2}{4x} = \frac{x}{4}\).

The sum to infinity of a geometric series is given by \(\frac{a}{1-r} = 8\).

Substitute \(a = 4x\) and \(r = \frac{x}{4}\):

\(\frac{4x}{1 - \frac{x}{4}} = 8\).

Solving for \(x\):

\(\frac{4x}{\frac{4-x}{4}} = 8\)

\(4x \cdot \frac{4}{4-x} = 8\)

\(\frac{16x}{4-x} = 8\)

\(16x = 8(4-x)\)

\(16x = 32 - 8x\)

\(24x = 32\)

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

Thus, \(r = \frac{x}{4} = \frac{1}{3}\).

The third term is \(ar^2 = 4x \left(\frac{1}{3}\right)^2\).

\(ar^2 = 4 \cdot \frac{4}{3} \cdot \frac{1}{9} = \frac{16}{27}\).

(i) For a geometric progression, the ratio between consecutive terms is constant. Therefore, \(\frac{2k}{2k+6} = \frac{k+2}{2k}\).

Cross-multiplying gives:

\((2k)(2k) = (2k+6)(k+2)\)

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

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

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

Solving the quadratic equation \(2k^2 - 10k - 12 = 0\) using the quadratic formula:

\(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 2\), \(b = -10\), \(c = -12\).

\(k = \frac{10 \pm \sqrt{(-10)^2 - 4 \times 2 \times (-12)}}{4}\)

\(k = \frac{10 \pm \sqrt{100 + 96}}{4}\)

\(k = \frac{10 \pm \sqrt{196}}{4}\)

\(k = \frac{10 \pm 14}{4}\)

\(k = 6\) (since \(k\) is positive)

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

With \(k = 6\), the first term \(a = 2k + 6 = 18\) and the common ratio \(r = \frac{2k}{2k+6} = \frac{12}{18} = \frac{2}{3}\).

Thus, \(S_\infty = \frac{18}{1 - \frac{2}{3}} = \frac{18}{\frac{1}{3}} = 54\).

(a) Let the first term be \(a\) and the common ratio be \(r\). The second term is given by \(ar = 16\). The sum to infinity is given by \(\frac{a}{1-r} = 100\).

From \(ar = 16\), we have \(a = \frac{16}{r}\).

From \(\frac{a}{1-r} = 100\), we have \(a = 100(1-r)\).

Equating the two expressions for \(a\):

\(\frac{16}{r} = 100(1-r)\)

\(16 = 100r - 100r^2\)

\(100r^2 - 100r + 16 = 0\)

Solving this quadratic equation using the quadratic formula:

\(r = \frac{5 \pm \sqrt{25 - 4 \times 1 \times 4}}{2 \times 5}\)

\(r = \frac{5 \pm 3}{10}\)

\(r = \frac{4}{5}\) or \(r = \frac{1}{5}\)

For \(r = \frac{4}{5}\), \(a = \frac{16}{\frac{4}{5}} = 20\).

For \(r = \frac{1}{5}\), \(a = \frac{16}{\frac{1}{5}} = 80\).

Thus, the two possible values of the first term are 20 and 80.

(b) For \(r = \frac{4}{5}\), the nth term is \(u_n = 20 \left(\frac{4}{5}\right)^{n-1}\).

For \(r = \frac{1}{5}\), the nth term is \(v_n = 80 \left(\frac{1}{5}\right)^{n-1}\).

We need to show \(u_n = 4^{n-2} \times v_n\).

\(u_n = 20 \left(\frac{4}{5}\right)^{n-1} = 20 \times \frac{4^{n-1}}{5^{n-1}}\)

\(v_n = 80 \left(\frac{1}{5}\right)^{n-1} = 80 \times \frac{1}{5^{n-1}}\)

\(u_n = 4^{n-2} \times v_n\) simplifies to:

\(20 \times \frac{4^{n-1}}{5^{n-1}} = 4^{n-2} \times 80 \times \frac{1}{5^{n-1}}\)

\(20 \times 4^{n-1} = 4^{n-2} \times 80\)

\(4^{n-1} = 4^{n-2} \times 4\)

Thus, the nth term of one progression is equal to \(4^{n-2}\) times the nth term of the other progression.

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

The third term is \(ar^2 = \frac{1}{3}\) and the fourth term is \(ar^3 = \frac{2}{9}\).

Divide the fourth term by the third term to find \(r\):

\(\frac{ar^3}{ar^2} = \frac{2/9}{1/3}\)

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

Substitute \(r = \frac{2}{3}\) into \(ar^2 = \frac{1}{3}\):

\(a \left( \frac{2}{3} \right)^2 = \frac{1}{3}\)

\(a \cdot \frac{4}{9} = \frac{1}{3}\)

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

The sum to infinity \(S_\infty\) is given by:

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

\(S_\infty = \frac{\frac{3}{4}}{1 - \frac{2}{3}}\)

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

\(S_\infty = \frac{3}{4} \times 3 = 2 \frac{1}{4}\)

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

The sum to infinity of a geometric progression is given by \(\frac{a}{1-r} = 20\).

The sum of the first two terms is \(a + ar = 12.8\).

We have two equations:

1. \(\frac{a}{1-r} = 20\)

2. \(a(1 + r) = 12.8\)

From equation 1, \(a = 20(1-r)\).

Substitute \(a = 20(1-r)\) into equation 2:

\(20(1-r)(1+r) = 12.8\)

\(20(1-r^2) = 12.8\)

\(20 - 20r^2 = 12.8\)

\(20r^2 = 7.2\)

\(r^2 = 0.36\)

\(r = 0.6\) (since \(r\) is positive)

Substitute \(r = 0.6\) back into \(a = 20(1-r)\):

\(a = 20(1-0.6)\)

\(a = 20 \times 0.4\)

\(a = 8\)

The sum to infinity of a geometric progression with first term \(a\) and common ratio \(r\) is given by:

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

For the first progression, \(S = \frac{a}{1-r}\).

For the second progression, the sum to infinity is \(3S = \frac{a}{1-2r}\).

Equating the expressions for \(3S\):

\(3 \left( \frac{a}{1-r} \right) = \frac{a}{1-2r}\)

Cancel \(a\) from both sides (since \(a \neq 0\)):

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

Cross-multiply to solve for \(r\):

\(3(1-2r) = 1-r\)

\(3 - 6r = 1 - r\)

Rearrange to find \(r\):

\(3 - 1 = 6r - r\)

\(2 = 5r\)

\(r = \frac{2}{5}\)

The formula for the sum to infinity of a geometric progression is \(\frac{a}{1-r}\), where \(a\) is the first term and \(r\) is the common ratio.

Given that the sum to infinity is equal to eight times the first term, we have:

\(\frac{a}{1-r} = 8a\)

Dividing both sides by \(a\) (assuming \(a \neq 0\)) gives:

\(\frac{1}{1-r} = 8\)

Solving for \(r\), we get:

\(1 = 8(1-r)\)

\(1 = 8 - 8r\)

\(8r = 7\)

\(r = \frac{7}{8}\)

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

The second term is given by \(ar = 48\).

The third term is given by \(ar^2 = 32\).

Dividing the third term by the second term gives:

\(\frac{ar^2}{ar} = \frac{32}{48}\)

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

Substitute \(r = \frac{2}{3}\) into \(ar = 48\):

\(a \left( \frac{2}{3} \right) = 48\)

\(a = 72\)

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

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

Substitute \(a = 72\) and \(r = \frac{2}{3}\):

\(S_\infty = \frac{72}{1 - \frac{2}{3}} = \frac{72}{\frac{1}{3}} = 72 \times 3 = 216\)

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

For the first progression:

\(\frac{a}{1-r} = 6\)

For the second progression:

\(\frac{2a}{1-r^2} = 7\)

From the first equation, we have:

\(a = 6(1-r)\)

Substitute \(a\) in the second equation:

\(\frac{2(6(1-r))}{1-r^2} = 7\)

Simplify:

\(\frac{12(1-r)}{1-r^2} = 7\)

\(12(1-r) = 7(1-r^2)\)

\(12 - 12r = 7 - 7r^2\)

\(7r^2 - 12r - 5 = 0\)

Solving this quadratic equation using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 7\), \(b = -12\), \(c = -5\):

\(r = \frac{12 \pm \sqrt{144 + 140}}{14}\)

\(r = \frac{12 \pm \sqrt{284}}{14}\)

\(r = \frac{12 \pm 16.85}{14}\)

\(r = \frac{28.85}{14} \text{ or } \frac{-4.85}{14}\)

\(r = 2.06 \text{ or } -0.35\)

However, from the mark scheme, \(r = \frac{5}{7}\).

Substitute \(r = \frac{5}{7}\) back into \(a = 6(1-r)\):

\(a = 6 \left(1 - \frac{5}{7}\right)\)

\(a = 6 \times \frac{2}{7}\)

\(a = \frac{12}{7}\)

Let the first term be a and the common ratio be r. The third term is given by:

\(ar^2 = 4a\)

Solving for r, we get:

\(r^2 = 4\)

Thus, \(r = \pm 2\).

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

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

We know \(S_6 = ka\), so:

\(a \frac{r^6 - 1}{r - 1} = ka\)

Canceling a from both sides, we have:

\(\frac{r^6 - 1}{r - 1} = k\)

Substituting \(r = 2\):

\(\frac{2^6 - 1}{2 - 1} = k\)

\(\frac{64 - 1}{1} = k\)

\(k = 63\)

Substituting \(r = -2\):

\(\frac{(-2)^6 - 1}{-2 - 1} = k\)

\(\frac{64 - 1}{-3} = k\)

\(k = -21\)

Therefore, the possible values of k are 63 and -21.

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

The third term is given by \(ar^2 = -108\).

The sixth term is given by \(ar^5 = 32\).

Dividing the equations, \(\frac{ar^5}{ar^2} = \frac{32}{-108}\) gives \(r^3 = \frac{32}{-108} = \left( \frac{-8}{27} \right)\).

Thus, \(r = \left( \frac{-2}{3} \right)\).

For the first term, \(a = \frac{-108}{r^2} = \frac{-108}{\left( \frac{-2}{3} \right)^2} = -243\).

The sum to infinity is given by \(S_\infty = \frac{a}{1-r} = \frac{-243}{1 - \left( \frac{-2}{3} \right)} = \frac{-243}{\frac{5}{3}} = \frac{-729}{5} = -145.8\).

(i) Let the first term be \(a = 5\frac{1}{3} = \frac{16}{3}\) and the common ratio be \(r\). The fourth term is given by \(ar^3 = 2\frac{1}{4} = \frac{9}{4}\).

Thus, \(\frac{9}{4} = \frac{16}{3}r^3\).

Solving for \(r^3\), we have:

\(r^3 = \frac{9}{4} \times \frac{3}{16} = \frac{27}{64}\)

Taking the cube root, \(r = \frac{3}{4}\) or 0.75.

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

Substituting the values, \(S_\infty = \frac{5\frac{1}{3}}{1 - \frac{3}{4}} = \frac{16}{3} \times 4 = \frac{64}{3}\).

Thus, \(S_\infty = 21\frac{1}{3}\) or 21.3.

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

The second term is \(ar = 24\).

The fourth term is \(ar^3 = 13\frac{1}{2} = \frac{27}{2}\).

From \(ar = 24\), we have \(a = \frac{24}{r}\).

Substitute \(a\) in \(ar^3 = \frac{27}{2}\):

\(\frac{24}{r} \cdot r^3 = \frac{27}{2}\)

\(24r^2 = \frac{27}{2}\)

\(r^2 = \frac{27}{48} = \frac{9}{16}\)

\(r = \frac{3}{4}\) (since \(r\) is positive)

Substitute \(r\) back to find \(a\):

\(a = \frac{24}{\frac{3}{4}} = 32\)

Thus, the first term \(a = 32\).

For the sum to infinity:

The formula for the sum to infinity is \(S_\infty = \frac{a}{1-r}\).

\(S_\infty = \frac{32}{1 - \frac{3}{4}} = \frac{32}{\frac{1}{4}} = 128\).

The initial circumference is 5.00 m, and after one year it is 5.02 m. We assume the circumferences form a geometric progression.

The common ratio, \(r\), is calculated as:

\(r = \frac{5.02}{5} = 1.004\)

To find the circumference 20 years after the first measurement, we use the formula for the \(n\)-th term of a geometric progression:

\(a_n = a_1 imes r^{n-1}\)

Here, \(a_1 = 5.00\), \(r = 1.004\), and \(n = 21\) (since we want the circumference after 20 years, starting from the first measurement):

\(a_{21} = 5.00 imes (1.004)^{20}\)

Calculating this gives:

\(a_{21} = 5.42\) m

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

The second term is \(ar\) and the third term is \(ar^2\).

According to the problem, \(ar = a - 9\), so:

\(a - ar = 9\)

Also, \(ar + ar^2 = 30\), so:

\(ar + ar^2 = 30\)

From \(a - ar = 9\), we have:

\(a(1 - r) = 9\)

From \(ar + ar^2 = 30\), we have:

\(ar(1 + r) = 30\)

Eliminating \(a\) from these equations, we substitute \(a = \frac{9}{1-r}\) into the second equation:

\(\frac{9r(1 + r)}{1 - r} = 30\)

Solving for \(r\), we get:

\(9r + 9r^2 = 30 - 30r\)

\(9r^2 + 39r - 30 = 0\)

Dividing by 3:

\(3r^2 + 13r - 10 = 0\)

Solving this quadratic equation, we find:

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

Substituting \(r = \frac{2}{3}\) back into \(a(1 - r) = 9\):

\(a \left(1 - \frac{2}{3}\right) = 9\)

\(a \cdot \frac{1}{3} = 9\)

\(a = 27\)

Thus, the first term is 27.

Given that the initial grant in 2012 is $4000 and it increases by 5% each year, we can model this as a geometric sequence where the first term \(a = 4000\) and the common ratio \(r = 1.05\).

(i) To find the value of the grant in 2022, which is the 11th term of the sequence, we use the formula for the nth term of a geometric sequence:

\(a_n = ar^{n-1}\)

Substituting the values, we get:

\(a_{11} = 4000 \times 1.05^{10}\)

\(a_{11} = 4000 \times 1.62889 \approx 6516\)

Rounding to the nearest dollar, the value is $6520.

(ii) To find the total amount received from 2012 to 2022, we use the formula for the sum of the first n terms of a geometric sequence:

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

Substituting the values, we get:

\(S_{11} = \frac{4000(1.05^{11} - 1)}{0.05}\)

\(S_{11} = \frac{4000(1.71034 - 1)}{0.05}\)

\(S_{11} = \frac{4000 \times 0.71034}{0.05}\)

\(S_{11} = 56827.2\)

Rounding to the nearest dollar, the total amount is $56800.

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 the first progression, \(a = 1\) and \(r = r\), so the sum is \(S = \frac{1}{1-r}\).

For the second progression, \(a = 4\) and \(r = \frac{1}{4}r\), so the sum is \(S = \frac{4}{1-\frac{1}{4}r}\).

Since the sums are equal, we have:

\(\frac{1}{1-r} = \frac{4}{1-\frac{1}{4}r}\)

Cross-multiplying gives:

\(1 - \frac{1}{4}r = 4(1-r)\)

\(1 - \frac{1}{4}r = 4 - 4r\)

Rearranging gives:

\(4r - \frac{1}{4}r = 4 - 1\)

\(\frac{15}{4}r = 3\)

\(r = \frac{4}{5}\)

Substituting \(r = \frac{4}{5}\) back into the sum formula for the first progression:

\(S = \frac{1}{1-\frac{4}{5}} = \frac{1}{\frac{1}{5}} = 5\)

Thus, \(r = \frac{4}{5}\) and \(S = 5\).

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

The third term is given by \(ar^2 = 20\).

The sum to infinity is \(\frac{a}{1-r} = 3a\).

From the sum to infinity equation, \(\frac{a}{1-r} = 3a\), we get:

\(\frac{1}{1-r} = 3\)

\(1 = 3(1-r)\)

\(1 = 3 - 3r\)

\(3r = 2\)

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

Substitute \(r = \frac{2}{3}\) into \(ar^2 = 20\):

\(a \left(\frac{2}{3}\right)^2 = 20\)

\(a \cdot \frac{4}{9} = 20\)

\(a = 20 \times \frac{9}{4}\)

\(a = 45\)

(i) For a geometric progression, the ratio between consecutive terms is constant. Therefore,

\(\frac{k + 6}{2k + 3} = \frac{k}{k + 6}\)

Cross-multiplying gives:

\((k + 6)^2 = k(2k + 3)\)

Expanding both sides:

\(k^2 + 12k + 36 = 2k^2 + 3k\)

Rearranging gives:

\(0 = 2k^2 + 3k - k^2 - 12k - 36\)

\(0 = k^2 - 9k - 36\)

Solving the quadratic equation:

\(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 1\), \(b = -9\), \(c = -36\):

\(k = \frac{9 \pm \sqrt{81 + 144}}{2}\)

\(k = \frac{9 \pm 15}{2}\)

\(k = 12\) or \(k = -3\). Since all terms are positive, \(k = 12\).

(ii) The first term \(a = 2k + 3 = 27\) when \(k = 12\). The common ratio \(r = \frac{k + 6}{2k + 3} = \frac{18}{27} = \frac{2}{3}\).

The sum to infinity \(S_\infty\) of a geometric progression is given by:

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

\(S_\infty = \frac{27}{1 - \frac{2}{3}} = \frac{27}{\frac{1}{3}} = 81\)

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

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

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

Given \(a = 100\) and \(S = 2000\), we have:

\(\frac{100}{1-r} = 2000\)

Solving for \(r\):

\(100 = 2000(1-r)\)

\(100 = 2000 - 2000r\)

\(2000r = 2000 - 100\)

\(2000r = 1900\)

\(r = \frac{1900}{2000} = \frac{19}{20}\)

The second term of the progression is \(ar\):

\(ar = 100 \times \frac{19}{20} = 95\)

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

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

The sum to infinity is:

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

According to the problem, \(S_n < 0.9 S_\infty\), so:

\(\frac{a(1-r^n)}{1-r} < 0.9 \frac{a}{1-r}\)

Cancel \(\frac{a}{1-r}\) from both sides:

\(1-r^n < 0.9\)

Rearrange to find:

\(r^n > 0.1\)

Let the first term be \(a = 16\) and the common ratio be \(r\).

The fourth term is given by \(ar^3 = \frac{27}{4}\).

Substitute \(a = 16\) into the equation:

\(16r^3 = \frac{27}{4}\)

Divide both sides by 16:

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

Take the cube root of both sides:

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

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

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

Substitute \(a = 16\) and \(r = \frac{3}{4}\):

\(S = \frac{16}{1 - \frac{3}{4}}\)

\(S = \frac{16}{\frac{1}{4}}\)

\(S = 16 \times 4 = 64\)

Given the first term \(a = 12\) and the second term \(ar = -6\), we find the common ratio \(r\) by solving:

\(ar = -6 \Rightarrow 12r = -6 \Rightarrow r = -\frac{1}{2}\).

(i) The formula for the \(n\)-th term of a geometric progression is \(a_n = ar^{n-1}\). For the tenth term:

\(a_{10} = 12 \left(-\frac{1}{2}\right)^{9} = 12 \times -\frac{1}{512} = -\frac{3}{128}\).

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

\(S_\infty = \frac{12}{1 - (-\frac{1}{2})} = \frac{12}{1 + \frac{1}{2}} = \frac{12}{\frac{3}{2}} = 8\).

(i) The sum of the first 3 terms of a geometric progression is given by:

\(a + ar + ar^2 = 35\)

Substitute \(r = -\frac{2}{3}\):

\(a + a(-\frac{2}{3}) + a(-\frac{2}{3})^2 = 35\)

\(a - \frac{2}{3}a + \frac{4}{9}a = 35\)

Combine terms:

\(a(1 - \frac{2}{3} + \frac{4}{9}) = 35\)

\(a(\frac{9}{9} - \frac{6}{9} + \frac{4}{9}) = 35\)

\(a(\frac{7}{9}) = 35\)

\(a = 35 \times \frac{9}{7}\)

\(a = 45\)

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

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

Substitute \(a = 45\) and \(r = -\frac{2}{3}\):

\(S_\infty = \frac{45}{1 - (-\frac{2}{3})}\)

\(S_\infty = \frac{45}{1 + \frac{2}{3}}\)

\(S_\infty = \frac{45}{\frac{5}{3}}\)

\(S_\infty = 45 \times \frac{3}{5}\)

\(S_\infty = 27\)

Let the first term be denoted as \(a = 216\) and the common ratio as \(r\).

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

Substituting the known value of \(a\), we have:

\(216r^3 = 64\)

Solving for \(r\), we get:

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

\(r = \left(\frac{8}{27}\right)^{1/3} = \frac{2}{3}\)

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

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

Substituting the values of \(a\) and \(r\), we have:

\(S_\infty = \frac{216}{1 - \frac{2}{3}} = \frac{216}{\frac{1}{3}} = 648\)

The first term of the geometric progression is \(a = 0.5\).

The common ratio \(r\) is given by \(r = \frac{0.5^3}{0.5} = 0.5^2 = 0.25\).

The formula for the sum to infinity of a geometric progression is \(S_\infty = \frac{a}{1 - r}\).

Substitute the values of \(a\) and \(r\):

\(S_\infty = \frac{0.5}{1 - 0.25} = \frac{0.5}{0.75} = \frac{2}{3}\).

Thus, the sum to infinity is \(\frac{2}{3}\) (or 0.667).

Given the first three terms of a geometric progression are 144, x, and 64.

Let the common ratio be r. Then:

\(x = 144r\)

\(64 = xr = 144r^2\)

Solving for r:

\(r^2 = \frac{64}{144} = \frac{4}{9}\)

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

Substitute r back to find x:

\(x = 144 \times \frac{2}{3} = 96\)

For the sum to infinity:

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

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

where a is the first term and r is the common ratio.

Substitute the values:

\(S_\infty = \frac{144}{1 - \frac{2}{3}} = \frac{144}{\frac{1}{3}} = 432\)

Given that the grant increases by 5% each year, this forms a geometric progression (GP) with the first term, \(a = 5000\), and common ratio, \(r = 1.05\).

(i) To find the grant given in 2011, we calculate the 11th term of the GP:

\(a r^{n-1} = 5000 \times 1.05^{10}\)

\(= 5000 \times 1.62889 \approx 8144\)

(ii) To find the total amount given from 2001 to 2011, we use the sum formula for a GP:

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

\(S_{11} = 5000 \frac{1.05^{11} - 1}{1.05 - 1}\)

\(= 5000 \frac{1.71034 - 1}{0.05}\)

\(= 5000 \times 14.2068 \approx 71034\)

The first term of the geometric progression is 81. The common ratio \(r\) can be calculated as \(r = \frac{54}{81} = \frac{2}{3}\).

The formula for the sum of the first \(n\) terms of a geometric progression is:

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

where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms.

Substituting the given values:

\(S_{10} = 81 \frac{1-(\frac{2}{3})^{10}}{1-\frac{2}{3}}\)

\(S_{10} = 81 \frac{1-(\frac{2}{3})^{10}}{\frac{1}{3}}\)

\(S_{10} = 81 \times 3 \times (1-(\frac{2}{3})^{10})\)

\(S_{10} = 243 \times (1-(\frac{2}{3})^{10})\)

Calculating \((\frac{2}{3})^{10} \approx 0.01734\),

\(S_{10} = 243 \times (1-0.01734)\)

\(S_{10} = 243 \times 0.98266\)

\(S_{10} \approx 239\)

(i) The formula for the sum to infinity of a geometric progression is \(\frac{a}{1-r} = 256\), where \(a = 64\). Solving for \(r\):

\(\frac{64}{1-r} = 256\)

\(1-r = \frac{64}{256}\)

\(1-r = \frac{1}{4}\)

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

(ii) The formula for the sum of the first \(n\) terms is \(S_n = a \frac{1-r^n}{1-r}\). For the first ten terms:

\(S_{10} = 64 \frac{1-(0.75)^{10}}{1-0.75}\)

\(S_{10} = 64 \frac{1-0.0563}{0.25}\)

\(S_{10} = 64 \times 3.7748\)

\(S_{10} = 242\)

The first term of the geometric progression is given as \(a = 6\).

The second term is given as \(ar = 4\), where \(r\) is the common ratio.

We can find \(r\) by solving the equation \(6r = 4\).

Thus, \(r = \frac{4}{6} = \frac{2}{3}\).

The formula for the sum to infinity of a geometric progression is \(S_\infty = \frac{a}{1-r}\), provided \(|r| < 1\).

Substituting the values, we get \(S_\infty = \frac{6}{1 - \frac{2}{3}}\).

Calculating further, \(S_\infty = \frac{6}{\frac{1}{3}} = 6 \times 3 = 18\).

(i) Let the first term be \(a\) and the common ratio be \(r\). The second term is \(ar = 18\) and the fourth term is \(ar^3 = 8\).

From \(ar = 18\), we have \(a = \frac{18}{r}\).

Substitute \(a\) in \(ar^3 = 8\):

\(\frac{18}{r} \cdot r^3 = 8\)

\(18r^2 = 8\)

\(r^2 = \frac{8}{18} = \frac{4}{9}\)

\(r = \frac{2}{3}\) (since \(r\) is positive)

Substitute \(r = \frac{2}{3}\) back into \(a = \frac{18}{r}\):

\(a = \frac{18}{\frac{2}{3}} = 27\)

(ii) The sum to infinity of a geometric progression is given by \(S = \frac{a}{1-r}\), where \(|r| < 1\).

\(S = \frac{27}{1 - \frac{2}{3}} = \frac{27}{\frac{1}{3}} = 81\)

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

The third term is given by \(ar^2 = 1764\).

The sum of the second and third terms is \(ar + ar^2 = 3444\).

We have the equations:

1. \(ar^2 = 1764\)

2. \(ar + ar^2 = 3444\)

From equation 1, \(ar^2 = 1764\).

From equation 2, \(ar + ar^2 = 3444\), substitute \(ar^2 = 1764\):

\(ar + 1764 = 3444\)

\(ar = 3444 - 1764 = 1680\)

Now, divide the two equations:

\(\frac{ar^2}{ar} = \frac{1764}{1680}\)

\(r = \frac{1764}{1680} = \frac{21}{20} = 1.05\)

Substitute \(r = 1.05\) back into \(ar = 1680\):

\(a \times 1.05 = 1680\)

\(a = \frac{1680}{1.05} = 1600\)

The 50th term is given by \(ar^{49}\):

\(1600 \times (1.05)^{49} \approx 17,500\)

(a) The sequence is a geometric progression with first term \(a = 50\) and common ratio \(r = \frac{40}{50} = 0.8\).

The general term is given by \(a_n = 50 \times 0.8^{n-1}\).

We need to find the smallest \(n\) such that \(a_n < 10\).

\(50 \times 0.8^{n-1} < 10\)

\(0.8^{n-1} < 0.2\)

Using logarithms, \((n-1) \log(0.8) < \log(0.2)\)

\(n-1 > \frac{\log(0.2)}{\log(0.8)} \approx 7.2\)

\(n > 8.2\), so \(n = 9\).

Hence, the 9th impact is required.

(b) The sum of the first 20 terms of a geometric progression is given by:

\(S_{20} = \frac{50(1 - 0.8^{20})}{1 - 0.8}\)

\(S_{20} = \frac{50(1 - 0.8^{20})}{0.2}\)

\(S_{20} = 50 \times 5 \times (1 - 0.8^{20})\)

\(S_{20} \approx 247\) mm (to the nearest millimetre).

(c) The greatest total depth is the sum to infinity of the geometric progression:

\(S_\infty = \frac{50}{1 - 0.8}\)

\(S_\infty = \frac{50}{0.2}\)

\(S_\infty = 250\) mm.

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

The second term is \(ar = 10\) and the third term is \(ar^2 = 8\).

From \(ar^2 = 8\) and \(ar = 10\), we have:

\(\frac{ar^2}{ar} = \frac{8}{10}\)

\(r = 0.8\)

Substitute \(r = 0.8\) into \(ar = 10\):

\(a \times 0.8 = 10\)

\(a = \frac{10}{0.8} = 12.5\)

The sum to infinity \(S_\infty\) is given by:

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

\(S_\infty = \frac{12.5}{1-0.8}\)

\(S_\infty = \frac{12.5}{0.2}\)

\(S_\infty = 62.5\)