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.
