[tex]S_n = \frac{n}{2}(a_1+a_n)[/tex]
[tex]a_1=16 [/tex] since it is the first multiple of 8 after 9.
[tex]a_n = 192[/tex] since it is the last multiple of 8 before 199.
n = 23 because there are a total of 23 multiples of 8 between 9 and 199.
[tex]S_{23} = \frac{23}{2}(16+192)=\frac{23}{2}(208)=23(104)=\boxed{2392}[/tex]
Therefore, the sum of all he integer multiples of 8 between 9 and 199 is 2,392.
