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

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

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

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