Since we are decreasing by 5 each time, we can build up a series formula first.
[tex]S_n = \frac{n}{2}(2a + (n - 1)d)[/tex]
[tex]S_n = \frac{n}{2}(200 - 5(n - 1))[/tex]
Now, adding summation notation, we produce:
[tex]\sum_{k = 1}^{n}[\frac{k}{2}(200 - 5(k - 1))][/tex]
