Let the first term be \(a = 32\) and the common difference be \(d\).

The 5th term is given by \(a + 4d = 22\).

Substituting \(a = 32\), we have:

\(32 + 4d = 22\)

\(4d = 22 - 32\)

\(4d = -10\)

\(d = -2.5\)

The last term is \(-28\), so:

\(a + (n-1)d = -28\)

\(32 + (n-1)(-2.5) = -28\)

\(32 - 2.5(n-1) = -28\)

\(32 - 2.5n + 2.5 = -28\)

\(34.5 - 2.5n = -28\)

\(-2.5n = -28 - 34.5\)

\(-2.5n = -62.5\)

\(n = 25\)

The sum of an arithmetic progression is given by:

\(S_n = \frac{n}{2} (a + l)\)

where \(l\) is the last term.

\(S_{25} = \frac{25}{2} (32 + (-28))\)

\(S_{25} = \frac{25}{2} (4)\)

\(S_{25} = 25 \times 2\)

\(S_{25} = 50\)

(i) The distance cycled on the nth day is given by the formula for the nth term of an arithmetic sequence:

\(a_n = a + (n-1) imes d\)

where \(a = 200\) km (the distance on the first day) and \(d = -5\) km (the reduction each day). For May 15th, \(n = 15\):

\(a_{15} = 200 + (15-1) imes (-5)\)

\(a_{15} = 200 - 70 = 130\) km

(ii) The total distance cycled is the sum of an arithmetic series:

\(S_n = \frac{n}{2} [2a + (n-1) imes d]\)

We need \(S_n = 3050\) km, \(a = 200\), and \(d = -5\):

\(\frac{n}{2} [400 + (n-1) imes (-5)] = 3050\)

\(\frac{n}{2} [400 - 5n + 5] = 3050\)

\(n(405 - 5n) = 6100\)

\(5n^2 - 405n + 6100 = 0\)

Solving this quadratic equation gives \(n = 20\).

Therefore, he finishes on May 20th.

(a) The amount of water lost each day forms an arithmetic progression (AP) with the first term, \(a = 10\), and the common difference, \(d = 2\).

The formula for the nth term of an AP is \(a + (n-1)d\).

For the 30th day, \(n = 30\), so the amount lost is:

\(10 + (30-1) imes 2 = 10 + 29 imes 2 = 10 + 58 = 68\) litres.

(b) The total amount of water lost by the nth day is given by the sum of the first n terms of the AP:

\(S_n = \frac{n}{2} (2a + (n-1)d)\).

We need \(S_n = 2000\):

\(\frac{n}{2} (20 + 2(n-1)) = 2000\)

\(n(20 + 2n - 2) = 4000\)

\(n(18 + 2n) = 4000\)

\(2n^2 + 18n - 4000 = 0\)

Solving this quadratic equation gives \(n = 41\).

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

The 12th term is given by:

\(a + 11d = 17\)

The sum of the first 31 terms is given by:

\(\frac{31}{2} (2a + 30d) = 1023\)

Solving the first equation for \(a\):

\(a = 17 - 11d\)

Substitute \(a\) in the second equation:

\(\frac{31}{2} (2(17 - 11d) + 30d) = 1023\)

\(\frac{31}{2} (34 - 22d + 30d) = 1023\)

\(\frac{31}{2} (34 + 8d) = 1023\)

\(31(34 + 8d) = 2046\)

\(34 + 8d = 66\)

\(8d = 32\)

\(d = 4\)

Substitute \(d = 4\) back into \(a = 17 - 11d\):

\(a = 17 - 44 = -27\)

The 31st term is given by:

\(a + 30d = -27 + 30 \times 4 = 93\)

Given that the progression is arithmetic, the common difference \(d\) is 12.

The first term is \(4x\) and the second term is \(x^2\).

Therefore, the common difference equation is:

\(x^2 - 4x = 12\)

Rearranging gives:

\(x^2 - 4x - 12 = 0\)

Solving this quadratic equation using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), \(c = -12\):

\(x = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}\)

\(x = \frac{4 \pm \sqrt{16 + 48}}{2}\)

\(x = \frac{4 \pm \sqrt{64}}{2}\)

\(x = \frac{4 \pm 8}{2}\)

\(x = 6\) or \(x = -2\)

For \(x = 6\), the third term is \(x^2 + 12 = 6^2 + 12 = 48\).

For \(x = -2\), the third term is \(x^2 + 12 = (-2)^2 + 12 = 16\).