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 $S_{40}=$ $\qquad$ .
9. 150 .
$$
\begin{array}{l}
\text { Let } b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, \\
b_{3}=S_{30}-S_{20}, b_{4}=S_{40}-S_{30} .
\end{array}
$$

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 common ratio $r=q^{10}$. Therefore,
$$
\begin{array}{l}
70=S_{30}=b_{1}+b_{2}+b_{3} \\
=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right) .
\end{array}
$$

Thus, $r^{2}+r-6=0$.
Solving, we get $r=2$ or $r=-3$ (discard).
Hence, $S_{40}=10\left(1+r+r^{2}+r^{3}\right)=150$.