Let's first show that $S_{10}$, $S_{20}-S_{10}$, $S_{30}-S_{20}$, $\cdots$ form a geometric sequence whose initial value is $S_{10}$ and common ratio is $r^{10}$ where $r$ is the common ratio of the original sequence $\{a_n\}$. \begin{align*} S_{10}&=a_1 \cdot \frac{1-r^{10}}{1-r}\\ S_{20}-S_{10}&=a_1 \cdot \frac{1-r^{20}}{1-r}-a_1\cdot \frac{1-r^{10}}{1-r}=a_1\cdot\frac{1-r^{10}}{1-}\cdot r^{10}\\ S_{30}-S_{20}&=a_1 \cdot \frac{1-r^{30}}{1-r}-a_1\cdot \frac{1-r^{20}}{1-r}=a_1\cdot\frac{1-r^{10}}{1-r}\cdot r^{20}\\ S_{40}-S_{30}&=a_1 \cdot \frac{1-r^{40}}{1-r}-a_1\cdot \frac{1-r^{30}}{1-r}=a_1\cdot\frac{1-r^{10}}{1-r}\cdot r^{30} \end{align*} Suppose $R=r^{10} \ge 0$, we will then have \begin{align*} S_30 &= S_{10} + (S_{20}-S{10})+(S_{30} - S_{20})\\ &= S_{10} + S_{10} \cdot R + S_{10} \cdot R^2\\ \therefore\quad 70 &= 10\times (1+R+R^2) \implies R&=2\\ \therefore\quad S_{40} &= S_{30} +(S_{40}-S_{30})\\ &= S_{30} + S_{10}\cdot R^3\\ &= 70 + 10 \times 2^3\\ &=\boxed{150} \end{align*}