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:

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

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