(i) The formula for the nth term of an arithmetic progression is given by:

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

For the ninth term, \(a + 8d = 22\).

The sum of the first 4 terms is given by:

\(S_4 = \frac{4}{2}(2a + 3d) = 49\)

\(2a + 3d = 24.5\)

We have the system of equations:

1. \(a + 8d = 22\)

2. \(2a + 3d = 24.5\)

Solving these equations simultaneously, we find:

\(d = 1.5\)

\(a = 10\)

(ii) The nth term is given by:

\(a + (n-1)d = 46\)

Substituting \(a = 10\) and \(d = 1.5\):

\(10 + (n-1) \times 1.5 = 46\)

\(1.5(n-1) = 36\)

\(n-1 = 24\)

\(n = 25\)

The first term of the arithmetic progression is given as \(a = 5\), and the second term is \(9\), so the common difference \(d = 9 - 5 = 4\).

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

We need to find the smallest \(n\) such that \(a_n > 200\).

Set \(a_n = 200\):

\(200 = 5 + (n-1) imes 4\)

\(200 = 5 + 4n - 4\)

\(200 = 4n + 1\)

\(199 = 4n\)

\(n = \frac{199}{4} = 49.75\)

Since \(n\) must be an integer, we take \(n = 50\) as the smallest integer greater than 49.75.

Now, calculate the sum of the first 50 terms using the sum formula for an arithmetic progression:

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

\(S_{50} = \frac{50}{2} (5 + 201)\)

\(S_{50} = 25 imes 206\)

\(S_{50} = 5150\)

The first term of the arithmetic progression is given as \(a = 6\).

The fifth term is given by the formula \(a + 4d = 12\).

Substituting \(a = 6\), we have:

\(6 + 4d = 12\)

\(4d = 6\)

\(d = 1.5\)

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

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

We know \(S_n = 90\), \(a = 6\), and \(d = 1.5\), so:

\(\frac{n}{2} (2 \times 6 + (n-1) \times 1.5) = 90\)

\(\frac{n}{2} (12 + 1.5n - 1.5) = 90\)

\(\frac{n}{2} (10.5 + 1.5n) = 90\)

\(n(10.5 + 1.5n) = 180\)

\(1.5n^2 + 10.5n - 180 = 0\)

Divide the entire equation by 1.5:

\(n^2 + 7n - 120 = 0\)

Solving this quadratic equation using the quadratic formula \(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 7\), \(c = -120\):

\(n = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-120)}}{2 \times 1}\)

\(n = \frac{-7 \pm \sqrt{49 + 480}}{2}\)

\(n = \frac{-7 \pm \sqrt{529}}{2}\)

\(n = \frac{-7 \pm 23}{2}\)

\(n = 8\) or \(n = -15\)

Since \(n\) must be positive, \(n = 8\).

To find the sum of all integers between 100 and 400 that are divisible by 7, we first identify the smallest and largest integers in this range that are divisible by 7.

The smallest integer greater than 100 that is divisible by 7 is 105, since 105 is the first multiple of 7 after 100.

The largest integer less than 400 that is divisible by 7 is 399, since 399 is the last multiple of 7 before 400.

These integers form an arithmetic sequence where the first term is 105, the last term is 399, and the common difference is 7.

The number of terms in this sequence, denoted as \(n\), can be found using the formula for the \(n\)-th term of an arithmetic sequence:

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

where \(l = 399\), \(a = 105\), and \(d = 7\).

Solving for \(n\):

\(399 = 105 + (n-1) imes 7\)

\(399 - 105 = (n-1) imes 7\)

\(294 = (n-1) imes 7\)

\(n-1 = \frac{294}{7}\)

\(n-1 = 42\)

\(n = 43\)

The sum of an arithmetic sequence is given by:

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

Substituting the known values:

\(S_{43} = \frac{43}{2} imes (105 + 399)\)

\(S_{43} = \frac{43}{2} imes 504\)

\(S_{43} = 43 imes 252\)

\(S_{43} = 10,836\)

Therefore, the sum of all integers between 100 and 400 that are divisible by 7 is 10,836.

The arithmetic progression (AP) is given by the sequence: 180, 175, 170, \ldots, 25.

The first term \(a = 180\) and the common difference \(d = 175 - 180 = -5\).

The last term \(l = 25\).

To find the number of terms \(n\), use the formula for the last term of an AP: \(l = a + (n-1) \cdot d\).

Substitute the known values: \(25 = 180 + (n-1) \cdot (-5)\).

Simplify: \(25 = 180 - 5(n-1)\).

\(25 = 180 - 5n + 5\).

\(25 = 185 - 5n\).

\(5n = 185 - 25\).

\(5n = 160\).

\(n = \frac{160}{5} = 32\).

Now, use the formula for the sum of an AP: \(S_n = \frac{n}{2} (a + l)\).

Substitute the known values: \(S_{32} = \frac{32}{2} (180 + 25)\).

\(S_{32} = 16 \cdot 205\).

\(S_{32} = 3280\).