The formula for the nth 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.

Given: \(a = 7\), \(a_n = 84\), and \(a_{3n} = 245\).

For the nth term:

\(7 + (n-1)d = 84\)

\((n-1)d = 77\)

For the (3n)th term:

\(7 + (3n-1)d = 245\)

\((3n-1)d = 238\)

Now, divide the two equations:

\(\frac{n-1}{3n-1} = \frac{77}{238}\)

Simplify \(\frac{77}{238}\) to \(\frac{11}{34}\):

\(\frac{n-1}{3n-1} = \frac{11}{34}\)

Cross-multiply:

\(34(n-1) = 11(3n-1)\)

\(34n - 34 = 33n - 11\)

\(n = 23\)

The sum of the first \(n\) terms of an arithmetic progression is given by:

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

For the first 200 terms:

\(S_{200} = \frac{200}{2} (2a + 199d) = 100(2a + 199d)\)

For the first 100 terms:

\(S_{100} = \frac{100}{2} (2a + 99d) = 50(2a + 99d)\)

Given that \(S_{200} = 4S_{100}\):

\(100(2a + 199d) = 4 \times 50(2a + 99d)\)

\(100(2a + 199d) = 200(2a + 99d)\)

\(2a + 199d = 2(2a + 99d)\)

\(2a + 199d = 4a + 198d\)

\(199d - 198d = 4a - 2a\)

\(d = 2a\)

For the 100th term:

The 100th term \(T_{100}\) is given by:

\(T_{100} = a + 99d\)

Substituting \(d = 2a\):

\(T_{100} = a + 99(2a)\)

\(T_{100} = a + 198a\)

\(T_{100} = 199a\)

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

The fifth term is given by:

\(a + 4d = 197\)

The sum of the first ten terms is given by:

\(\frac{10}{2} [2a + 9d] = 2040\)

Simplifying the sum equation:

\(5(2a + 9d) = 2040\)

\(2a + 9d = 408\)

We now have two equations:

1. \(a + 4d = 197\)

2. \(2a + 9d = 408\)

Multiply the first equation by 2:

\(2a + 8d = 394\)

Subtract this from the second equation:

\((2a + 9d) - (2a + 8d) = 408 - 394\)

\(d = 14\)

Thus, the common difference is 14.

(i) The time taken for each mile forms an arithmetic sequence where the first term is 5 minutes (or 300 seconds) and the common difference is 12 seconds. The nth term of an arithmetic sequence is given by:

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

where \(a = 300\) and \(d = 12\). We need to find \(n\) such that \(a_n = 540\) (since 9 minutes is 540 seconds):

\(540 = 300 + (n-1) imes 12\)

\(240 = (n-1) imes 12\)

\(n-1 = 20\)

\(n = 21\)

(ii) The total time for 26 miles is the sum of the arithmetic sequence with 26 terms. The sum \(S_n\) of an arithmetic sequence is given by:

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

For \(n = 26\), \(a = 300\), and \(d = 12\):

\(S_{26} = rac{26}{2} imes (2 imes 300 + 25 imes 12)\)

\(S_{26} = 13 imes (600 + 300)\)

\(S_{26} = 13 imes 900 = 11700\) seconds

Convert 11700 seconds to hours and minutes:

\(11700 ext{ seconds} = 3 ext{ hours and } 15 ext{ minutes}\)

The sum of the first ten terms of an arithmetic progression is given by:

\(S_{10} = \frac{10}{2} (2a + 9d) = 400\)

This simplifies to:

\(2a + 9d = 80\)

The sum of the next ten terms (terms 11 to 20) is given by:

\(S_{20} - S_{10} = 1000\)

\(\frac{20}{2} (2a + 19d) - \frac{10}{2} (2a + 9d) = 1000\)

\(2a + 19d = 140\)

We now have two equations:

1. \(2a + 9d = 80\)

2. \(2a + 19d = 140\)

Subtract equation 1 from equation 2:

\((2a + 19d) - (2a + 9d) = 140 - 80\)

\(10d = 60\)

\(d = 6\)

Substitute \(d = 6\) back into equation 1:

\(2a + 9(6) = 80\)

\(2a + 54 = 80\)

\(2a = 26\)

\(a = 13\)