First we will find the number of integers between 301 and 400 that are multiples of 4. We can see that the first such number is 304 being a multiple of 4. So the required numbers are 304, 308, 312,...,400 which form an AP.
To find the number of such integers use,
[tex] a_n=a_1+(n-1)d [/tex]. Here [tex] a_n [/tex] is the nth term, [tex] d [/tex] is the common difference of the AP. Here its is [tex] d=4 [/tex]. So
[tex] 400=304+(n-1)4\\
(n-1)4=96\\
n=25 [/tex]
The required sum of the AP is
[tex] S_n=\frac{n}{2} (a_1+a_n)\\
S_n=\frac{25}{2} (304+400)\\
S_n=8,800 [/tex]
