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.