$10 \cdot 38$ In a geometric sequence $\left\{a_{n}\right\}$ with all terms being real numbers, the sum of the first $n$ terms is denoted as $S_{n}$. If $S_{10}$ $=10, S_{30}=70$, then $S_{40}$ equals
(A) 150.
(B) -200.
(C) 150 or -200.
(D) 400 or -50.
(China High School Mathematics League, 1998)
[Solution] 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 $r=q^{10}$.
Thus, $70=S_{30}=b_{1}+b_{2}+b_{3}=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right)$.
$\therefore r^{2}+r-6=0$, solving this gives $r=2\left(\because r=q^{10}>0, \therefore r=-3\right.$ should be discarded).
Therefore, $S_{40}=10\left(1+2+2^{2}+2^{3}\right)=150$.
Hence, the answer is $(A)$.