Sequences

$S_n=\frac{n}{2}(a_1+(n-1)d)$
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:
$S_{70}=\frac{70}{2}(a_1+69d)\\ \boxed{S_{70}=\frac{21}{5}}$