Let the first term of the arithmetic progression be \(a = 60\) and the common difference be \(d\). The total number of payments is \(n = 48\), and the total amount repaid is \(S_n = 3726\).

The formula for the sum of an arithmetic progression is:

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

Substitute the given values:

\(3726 = \frac{48}{2} (2 \times 60 + 47d)\)

\(3726 = 24 (120 + 47d)\)

\(3726 = 2880 + 1128d\)

\(846 = 1128d\)

\(d = \frac{846}{1128} = 0.75\)

The third payment is given by the formula for the nth term of an arithmetic progression:

\(a_3 = a + 2d\)

\(a_3 = 60 + 2 \times 0.75\)

\(a_3 = 60 + 1.5\)

\(a_3 = 61.50\)

Let the first term be \(a = -10\) and the common difference be \(d\).

The 15th term is given by \(a + 14d = 11\).

Substituting \(a = -10\), we have:

\(-10 + 14d = 11\)

\(14d = 21\)

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

The last term is given by \(a + (n-1)d = 41\).

Substituting \(a = -10\) and \(d = \frac{3}{2}\), we have:

\(-10 + (n-1) \times \frac{3}{2} = 41\)

\((n-1) \times \frac{3}{2} = 51\)

\(n-1 = 34\)

\(n = 35\)

The sum of the arithmetic progression is given by:

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

\(S_{35} = \frac{35}{2} (-10 + 41)\)

\(S_{35} = \frac{35}{2} \times 31\)

\(S_{35} = 542.5\)

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

The formula for the \(n\)-th term of an arithmetic progression is \(a_n = a + (n-1)d\).

Given the 13th term is 12, we have:

\(a + 12d = 12\) (Equation 1)

The formula for the sum of the first \(n\) terms is \(S_n = \frac{n}{2} (2a + (n-1)d)\).

Given the sum of the first 30 terms is -15, we have:

\(\frac{30}{2} (2a + 29d) = -15\)

\(15(2a + 29d) = -15\)

\(2a + 29d = -1\) (Equation 2)

Solving Equations 1 and 2 simultaneously:

From Equation 1: \(a = 12 - 12d\)

Substitute into Equation 2:

\(2(12 - 12d) + 29d = -1\)

\(24 - 24d + 29d = -1\)

\(5d = -25\)

\(d = -5\)

Substitute \(d = -5\) back into Equation 1:

\(a + 12(-5) = 12\)

\(a - 60 = 12\)

\(a = 72\)

Now, find the sum of the first 50 terms:

\(S_{50} = \frac{50}{2} (2a + 49d)\)

\(S_{50} = 25 (2(72) + 49(-5))\)

\(S_{50} = 25 (144 - 245)\)

\(S_{50} = 25 (-101)\)

\(S_{50} = -2525\)

(a) The \(n\)th term of an arithmetic progression is given by \(a_n = a + (n-1)d\), where \(a\) is the first term and \(d\) is the common difference.

Here, \(a = 84\) and \(d = -3\). We need \(a_n < 0\).

\(84 + (n-1)(-3) < 0\)

\(84 - 3(n-1) < 0\)

\(84 - 3n + 3 < 0\)

\(87 < 3n\)

\(n > 29\)

The smallest integer \(n\) is 30.

(b) The sum of the first \(m\) terms of an arithmetic progression is given by \(S_m = \frac{m}{2} [2a + (m-1)d]\).

We are given \(S_{2k} = S_k\).

\(\frac{2k}{2} [2 \times 84 + (2k-1)(-3)] = \frac{k}{2} [2 \times 84 + (k-1)(-3)]\)

\(k [168 + (2k-1)(-3)] = \frac{k}{2} [168 + (k-1)(-3)]\)

\(k [168 - 6k + 3] = \frac{k}{2} [168 - 3k + 3]\)

\(k [171 - 6k] = \frac{k}{2} [171 - 3k]\)

\(2k [171 - 6k] = k [171 - 3k]\)

\(342k - 12k^2 = 171k - 3k^2\)

\(342k - 171k = 12k^2 - 3k^2\)

\(171k = 9k^2\)

\(19k = k^2\)

\(k = 19\)

For the arithmetic progression P, the 5th term is given by:

\(a + 4d.\)

For the arithmetic progression Q, the 12th term is given by:

\(2(a + 1) + 11(d + 1).\)

Using the given ratio:

\(\frac{a + 4d}{2(a + 1) + 11(d + 1)} = \frac{1}{3}.\)

Cross-multiplying gives:

\(3(a + 4d) = 2(a + 1) + 11(d + 1).\)

Simplifying, we get:

\(a + d = 13.\)

For the sum of the first 5 terms of P:

\(\frac{5}{2}(2a + 4d).\)

For the sum of the first 5 terms of Q:

\(\frac{5}{2}(4(a + 1) + 4(d + 1)).\)

Using the given ratio:

\(\frac{\frac{5}{2}(2a + 4d)}{\frac{5}{2}(4(a + 1) + 4(d + 1))} = \frac{2}{3}.\)

Cross-multiplying gives:

\(3(2a + 4d) = 2(4(a + 1) + 4(d + 1)).\)

Simplifying, we get:

\(-a + 2d = 8.\)

Solving the equations \(a + d = 13\) and \(-a + 2d = 8\) simultaneously, we find:

\(d = 7\) and \(a = 6\).