+1768 Let $a_1,$ $a_2,$ $a_3,$ $\dots$ be an arithmetic sequence. Let $S_n$ denote the sum of the first $n$ terms. If $S_{20} = \frac{1}{5}$ and $S_{10} = 0,$ then find $S_{70}.$
0 \(S_n=\frac{n}{2}(a_1+(n-1)d)\)
So using this and plugging in values we get:
\(\frac{1}{5}=\frac{20}{2}(a_1+19d)\\ 0=\frac{10}{2}(a_1+9d)\) solving we get:
\(a=-\frac{9}{500}, d=\frac{1}{500}\)
Therefore:
\(S_{70}=\frac{70}{2}(a_1+69d)\\ \boxed{S_{70}=\frac{21}{5}}\)
