(i) For an arithmetic progression, the second term is given by:

\(a + d = 96\)

and the fourth term is:

\(a + 3d = 54\)

Subtract the first equation from the second:

\((a + 3d) - (a + d) = 54 - 96\)

\(2d = -42\)

\(d = -21\)

Substitute \(d = -21\) back into \(a + d = 96\):

\(a - 21 = 96\)

\(a = 117\)

(ii) For a geometric progression, the second term is:

\(ar = 96\)

and the fourth term is:

\(ar^3 = 54\)

Divide the second equation by the first:

\(\frac{ar^3}{ar} = \frac{54}{96}\)

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

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

Substitute \(r = \frac{3}{4}\) back into \(ar = 96\):

\(a \times \frac{3}{4} = 96\)

\(a = 128\)

Let the first term of the geometric progression be 216 and the common ratio be \(r\).

The fourth term is given by \(216r^3 = 64\).

Solving for \(r\), we have:

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

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

The second term of the geometric progression is \(216r = 144\).

The third term of the geometric progression is \(216r^2 = 96\).

Let the first term of the arithmetic progression be \(a\) and the common difference be \(d\).

The second term of the arithmetic progression is \(a + d = 144\).

The fifth term of the arithmetic progression is \(a + 4d = 96\).

Solving the equations simultaneously:

\(a + d = 144\)

\(a + 4d = 96\)

Subtract the first equation from the second:

\(3d = -48\)

\(d = -16\)

Substitute \(d = -16\) into \(a + d = 144\):

\(a - 16 = 144\)

\(a = 160\)

The sum of the first 21 terms of the arithmetic progression is given by:

\(S_{21} = \frac{21}{2} (2a + 20d)\)

\(S_{21} = \frac{21}{2} (320 + 20(-16))\)

\(S_{21} = \frac{21}{2} (320 - 320)\)

\(S_{21} = 0\)

(i) The first term of the arithmetic progression is 8, so the first term of the geometric progression is also 8. The fifth term of the arithmetic progression is \(8 + 4d\), which is the second term of the geometric progression, so \(8 + 4d = 8r\). The eighth term of the arithmetic progression is \(8 + 7d\), which is the third term of the geometric progression, so \(8 + 7d = 8r^2\).

We have the equations:

\(8 + 4d = 8r\)

\(8 + 7d = 8r^2\)

Eliminating one of the variables, we solve:

\(4r^2 - 7r + 3 = 0\)

Solving this quadratic equation gives \(r = \frac{3}{4}\) and \(d = -\frac{1}{2}\).

(ii) The sum to infinity of the geometric progression is given by \(S_\infty = \frac{a}{1-r}\), where \(a = 8\) and \(r = \frac{3}{4}\).

\(S_\infty = \frac{8}{1 - \frac{3}{4}} = 32\)

(iii) The sum of the first 8 terms of the arithmetic progression is given by \(S_8 = \frac{n}{2} (2a + (n-1)d)\), where \(n = 8\), \(a = 8\), and \(d = -\frac{1}{2}\).

\(S_8 = 4(16 + 7(-\frac{1}{2})) = 50\)

(i) Let the first term be \(a = 81\) and the common ratio be \(r\). The fourth term is given by \(ar^3 = 24\).

Thus, \(81r^3 = 24\) which gives \(r^3 = \frac{24}{81}\).

Solving for \(r\), we get \(r = \frac{2}{3}\) or 0.667.

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

Substituting \(a = 81\) and \(r = \frac{2}{3}\), we get \(S_\infty = \frac{81}{1 - \frac{2}{3}} = 243\).

(iii) The second term of the GP is \(ar = 81 \times \frac{2}{3} = 54\).

The third term of the GP is \(ar^2 = 36\).

These are the first and fourth terms of the AP, so \(3d = -18\) giving \(d = -6\).

The sum of the first ten terms of the AP is \(S_{10} = 5 \times (108 - 54) = 270\).

(i) The general term of an arithmetic progression is given by \(a_n = a + (n-1)d\).

For the 5th term, \(n = 5\):

\(a_5 = a + 4d\).

For the 15th term, \(n = 15\):

\(a_{15} = a + 14d\).

(ii) The terms \(a\), \(a + 4d\), and \(a + 14d\) form a geometric progression.

Thus, \(\frac{a + 4d}{a} = \frac{a + 14d}{a + 4d}\).

Cross-multiplying gives:

\((a + 4d)^2 = a(a + 14d)\).

Expanding both sides:

\(a^2 + 8ad + 16d^2 = a^2 + 14ad\).

Simplifying gives:

\(8ad + 16d^2 = 14ad\).

\(16d^2 = 6ad\).

\(8d = 3a\).

Thus, \(3a = 8d\).

(iii) The common ratio \(r\) of the geometric progression is:

\(r = \frac{a + 4d}{a}\) or \(r = \frac{a + 14d}{a + 4d}\).

Using \(r = \frac{a + 4d}{a}\):

\(r = \frac{a + 4d}{a} = 1 + \frac{4d}{a}\).

Substituting \(3a = 8d\), we have \(a = \frac{8d}{3}\).

\(r = 1 + \frac{4d}{\frac{8d}{3}} = 1 + \frac{12d}{8d} = 1 + \frac{3}{2} = 2.5\).