Question 842311
Curiously even though we are talking about multiples of 1/5 it will be  an arithmetic sequence not geometric.
1/5, 2/5, 3/5,4/5


First term is 1/5 and the difference is 1/5
S = ½(2a + (n-1)d)n
positive S = ½(2/5 + (n-1)/5)n
negative S = ½(-2/5  -(n-1)/5)n
If we also include negative multiples of 1/5 
First term is -1/5 and the difference is -1/5
then the sum will be zero because the pluses and minuses add up to zero.

S = ½(2/5 + (n-1)/5)n+ ½(-2/5  -(n-1)/5)n=0