(9) Let the sum of the first $n$ terms of the real geometric sequence $\left\{a_{n}\right\}$ be $S_{n}$. If $S_{10}=10, S_{30}=$ 70, then $S_{40}=$ $\qquad$
9150 Prompt: Let $b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, b_{3}=S_{30}-S_{20}, b_{4}=$ $S_{40}-S_{30}$.

Let $q$ be 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 $r=q^{10}$, thus
$$
70=S_{30}=b_{1}+b_{2}+b_{3}=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right)
$$

which leads to $r^{2}+r-6=0$, solving this gives $r=2$ or $r=-3$ (discard), hence
$$
S_{40}=10\left(1+r+r^{2}+r^{3}\right)=150 \text {. }
$$