(i) The third term is given by \(a + 2d = 90\) and the fifth term by \(a + 4d = 80\). Solving these equations simultaneously:

Subtract the first from the second: \((a + 4d) - (a + 2d) = 80 - 90\)

\(2d = -10\)

\(d = -5\)

Substitute \(d = -5\) into \(a + 2d = 90\):

\(a + 2(-5) = 90\)

\(a - 10 = 90\)

\(a = 100\)

(ii) The sum of the first \(m\) terms is equal to the sum of the first \((m + 1)\) terms:

\(S_m = S_{m+1}\)

\(\frac{m}{2}(2a + (m-1)d) = \frac{m+1}{2}(2a + md)\)

Since \(S_m = S_{m+1}\), the difference is zero:

\(a + md = 0\)

\(100 + m(-5) = 0\)

\(100 - 5m = 0\)

\(5m = 100\)

\(m = 20\)

(iii) The sum of the first \(n\) terms is zero:

\(S_n = 0\)

\(\frac{n}{2}(2a + (n-1)d) = 0\)

\(n(200 + (n-1)(-5)) = 0\)

\(n(200 - 5n + 5) = 0\)

\(n(205 - 5n) = 0\)

\(205 - 5n = 0\)

\(5n = 205\)

\(n = 41\)

The first term of the arithmetic progression is 161, and the second term is 154. The common difference, d, is given by:

\(d = 154 - 161 = -7\)

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

\(S_m = \frac{m}{2} (2a + (m-1)d)\)

Substituting the given values, we have:

\(S_m = \frac{m}{2} (2 \times 161 + (m-1)(-7)) = 0\)

Simplifying inside the brackets:

\(2 \times 161 = 322\)

Thus, the equation becomes:

\(\frac{m}{2} (322 + (m-1)(-7)) = 0\)

Solving for m:

\(322 + (m-1)(-7) = 0\)

\(322 - 7m + 7 = 0\)

\(329 - 7m = 0\)

\(7m = 329\)

\(m = \frac{329}{7} = 47\)

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

The fifth term of the arithmetic progression is given by:

\(a + 4d = 18\)

The sum of the first 5 terms is given by:

\(\frac{5}{2} (2a + 4d) = 75\)

Solving the first equation for \(a\):

\(a = 18 - 4d\)

Substitute \(a = 18 - 4d\) into the sum equation:

\(\frac{5}{2} (2(18 - 4d) + 4d) = 75\)

\(\frac{5}{2} (36 - 8d + 4d) = 75\)

\(\frac{5}{2} (36 - 4d) = 75\)

\(5(36 - 4d) = 150\)

\(180 - 20d = 150\)

\(30 = 20d\)

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

Substitute \(d = \frac{3}{2}\) back into \(a = 18 - 4d\):

\(a = 18 - 4 \times \frac{3}{2}\)

\(a = 18 - 6\)

\(a = 12\)

Thus, the first term is 12, and the common difference is \(\frac{3}{2}\).

The multiples of 5 between 100 and 300 form an arithmetic sequence where the first term is 100 and the common difference is 5.

The sequence is: 100, 105, 110, ..., 300.

To find the number of terms, use the formula for the nth term of an arithmetic sequence:

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

where \(a = 100\), \(d = 5\), and \(a_n = 300\).

Solving for \(n\):

\(300 = 100 + (n-1) imes 5\)

\(300 = 100 + 5n - 5\)

\(300 = 95 + 5n\)

\(205 = 5n\)

\(n = 41\)

Now, use the formula for the sum of an arithmetic sequence:

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

Substitute the known values:

\(S_{41} = \frac{41}{2} imes (100 + 300)\)

\(S_{41} = \frac{41}{2} imes 400\)

\(S_{41} = 41 imes 200\)

\(S_{41} = 8200\)

Therefore, the sum of all the multiples of 5 between 100 and 300 inclusive is 8200.

(a) For an arithmetic progression, the difference between consecutive terms is constant. Therefore, we have:

\(6k - k = k + 6 - 6k\)

\(5k = k + 6 - 6k\)

\(5k = k + 6 - 6k\)

\(5k + 6k = k + 6\)

\(11k = k + 6\)

\(10k = 6\)

\(k = \frac{6}{10} = 0.6\)

(b) The common difference \(d\) is \(6k - k = 5k\). Substituting \(k = 0.6\), we get \(d = 3\).

The sum of the first 30 terms \(S_{30}\) is given by:

\(S_{30} = \frac{30}{2} (2 \times 0.6 + 29 \times 3)\)

\(S_{30} = 15 (1.2 + 87)\)

\(S_{30} = 15 \times 88.2\)

\(S_{30} = 1323\)