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