I use the function Drop quite often. When looking at this question, my attention was caught by a use of Drop in a way I did not see before. So I read the documentation and found the following description:
Drop[list,{m,n,s}] gives list with elements m through n in steps of s dropped.
As I read it, this formulation suggests that the list is partioned in sublists of length s and that in each of these sublists the elements m through n are dropped.
Drop[Range[10], {2,3,5}] => {1,3,4,5,6,7,8,9,10} I expected that the elements 2, 3, 7 and 8 would be dropped. It is only element 2 that is dropped. I get the same result when I replace the number 3 (n) with any number between 2 and 10. So with these values, m through n seems to result in only m for the original list.
In the next command I indeed see some periodictity with respect to s:
Drop[Range[10], {3,-1, 5}] => {1, 2, 4, 5, 6, 7, 9, 10} But I fail to see why elements 3 to -1 in steps of 5 results in 3 and 8. I get the same result when I replace -1 with -2 or -3.
Finally:
Drop[Range[10], {5,-1, 5}] => {1,2,3,4,6,7,8,9,10} Here elements 5 and 10 are dropped. But:
Drop[Range[10], {5,-2, 5}] => {1,2,3,4,6,7,8,9,10} Now only element 5 is dropped.
Can someone give me a better understanding how I can predict the result of Drop, when used with a list {m, n, s}?

m < n(ands > 0by necessity),Drop[list, {m, n, s}]dropsm + s*ifor allssuch thatm + s*i <= n. Ifn > mwe must have thats < 0and it will dropm + s*ifor allssuch thatm + s*i >= n. $\endgroup$