The problem describes a geometric series where the first term is the amount allocated in 2005, which is $2000, and the common ratio is the increase of 2.5% each year. Therefore, the common ratio is:

\(r = 1 + \frac{2.5}{100} = 1.025\)

The total amount allocated from 2005 to 2014 is the sum of the first 10 terms of this geometric series. The formula for the sum of the first \(n\) terms of a geometric series is:

\(S_n = a \frac{r^n - 1}{r - 1}\)

Substituting the known values \(a = 2000\), \(r = 1.025\), and \(n = 10\):

\(S_{10} = 2000 \frac{1.025^{10} - 1}{1.025 - 1}\)

Calculating this gives:

\(S_{10} = 2000 \frac{1.2800845 - 1}{0.025} \approx 22400\)

Therefore, the total amount allocated from 2005 to 2014 is $22,400.

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 progression P, the first term \(a = 6\) and the second term is \(6r\), so the common ratio \(r\) is \(r\).

The sum to infinity for P is \(S_P = \frac{6}{1 - r}\).

For progression Q, the first term \(a = 12\) and the second term is \(-12r\), so the common ratio \(r = -r\).

The sum to infinity for Q is \(S_Q = \frac{12}{1 + r}\).

Since the sums are equal, \(\frac{6}{1 - r} = \frac{12}{1 + r}\).

Cross-multiplying gives:

\(6(1 + r) = 12(1 - r)\)

\(6 + 6r = 12 - 12r\)

\(18r = 6\)

\(r = \frac{1}{3}\)

Substitute \(r = \frac{1}{3}\) back into the sum formula for P:

\(S = \frac{6}{1 - \frac{1}{3}} = \frac{6}{\frac{2}{3}} = 9\)

Let the first term be \(a\) and the common ratio be \(r\).

The third term is \(ar^2\) and the sixth term is \(ar^5\).

Given that \(ar^2 = 8 \times ar^5\), we have:

\(ar^2 = 8ar^5\)

\(r^2 = 8r^5\)

\(r^3 = \frac{1}{8}\)

\(r = \frac{1}{2}\)

The sum of the first six terms is given by:

\(S_6 = \frac{a(1-r^6)}{1-r} = 31.5\)

\(\frac{a(1-(\frac{1}{2})^6)}{\frac{1}{2}} = 31.5\)

\(\frac{a(1-\frac{1}{64})}{\frac{1}{2}} = 31.5\)

\(\frac{a(\frac{63}{64})}{\frac{1}{2}} = 31.5\)

\(\frac{63a}{64} \times 2 = 31.5\)

\(\frac{63a}{32} = 31.5\)

\(63a = 31.5 \times 32\)

\(63a = 1008\)

\(a = 16\)

The sum to infinity is given by:

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

\(S_\infty = \frac{16}{1-\frac{1}{2}}\)

\(S_\infty = \frac{16}{\frac{1}{2}}\)

\(S_\infty = 32\)

Let the first term be \(a\) and the common ratio be \(r\).

The first term is \(a\), the second term is \(ar\), and the third term is \(ar^2\).

According to the problem, we have:

\(a + ar = 50\)

\(ar + ar^2 = 30\)

From the first equation, factor out \(a\):

\(a(1 + r) = 50\)

From the second equation, factor out \(ar\):

\(ar(1 + r) = 30\)

Divide the second equation by the first equation:

\(\frac{ar(1 + r)}{a(1 + r)} = \frac{30}{50}\)

\(r = \frac{3}{5}\)

Substitute \(r = \frac{3}{5}\) back into the first equation:

\(a(1 + \frac{3}{5}) = 50\)

\(a(\frac{8}{5}) = 50\)

\(a = \frac{125}{4}\)

The sum to infinity \(S\) is given by:

\(S = \frac{a}{1 - r}\)

\(S = \frac{\frac{125}{4}}{1 - \frac{3}{5}}\)

\(S = \frac{\frac{125}{4}}{\frac{2}{5}}\)

\(S = \frac{125}{4} \times \frac{5}{2}\)

\(S = \frac{625}{8}\)

The amount of water lost each day forms a geometric progression (GP) with the first term, \(a = 10\), and common ratio, \(r = 1.1\).

The sum of the first 30 terms of a GP is given by:

\(S_{30} = \frac{a(r^{30} - 1)}{r - 1}\)

Substituting the values, we have:

\(S_{30} = \frac{10(1.1^{30} - 1)}{1.1 - 1}\)

Calculating this gives \(S_{30} = 1645\).

The total water lost after 30 days is 1645 litres.

The percentage of the original 2000 litres left in the tank is:

\(\text{Percentage left} = \frac{2000 - 1645}{2000} \times 100\)

\(= 17.75\%\)