Algebra

Let $d = a_2 -a_1$. Note that $S_5 = 5a_1 + (d + 2d + 3d + 4d) = 5a_1 + 10d$, and $S_{10} = 10a_1 + (d + 2d + \cdots + 9d) = 10a_1 + 45d$, and $S_{15} = 15a_1 + (d + 2d + \cdots + 14d) = 15a_1 + 105d$.

So, the given conditions are $\begin{cases} 5a_1 + 10d = \dfrac15\\ 10a_1 + 45d = \dfrac1{10} \end{cases}$. Solving gives $d = -\dfrac3{250}$ and $a_1 = \dfrac8{125}$.

So $S_{15} = 15a_1 + 105d = -\dfrac3{10}$.