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:

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

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

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

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