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.
