3.1.5 * For a geometric sequence $\left\{a_{n}\right\}$ with all terms being real numbers, the sum of the first $n$ terms is $S_{n}$. If $S_{10}=10$, $S_{30}=70$. Find $S_{40}$.
Let $b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, b_{3}=S_{30}-S_{20}, b_{4}=S_{40}-S_{30}$. Suppose $q$ is the common ratio of $\left\{a_{n}\right\}$, then $b_{1}, b_{2}, b_{3}, b_{4}$ form a geometric sequence with the common ratio $r=q^{10}$. Therefore,
$$
70=S_{30}=b_{1}+b_{2}+b_{3}=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right),
$$

which gives $r^{2}+r-6=0$, solving this yields $r=2$ or $r=-3$.
Since $r=q^{10}>0$, we have $r=2$, thus $S_{40}=10\left(1+2+2^{2}+2^{3}\right)=150$.