Example 3 (1998 National High School Competition Question) For the geometric sequence $\left\{a_{n}\right\}$ with real number terms, the sum of the first $n$ terms is $S_{n}$. If $S_{10}=10, S_{30}=70$, then $S_{40}$ equals ( ).
A. 150
B. -200
C. 150 or -200
D. 400 or -50
Solution: Choose A. Reason: Let $b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, b_{3}=S_{30}-S_{20}, b_{4}=S_{40}-S_{30}$, and denote the common ratio of $\left\{a_{n}\right\}$ as $q$, then $b_{1}, b_{2}, b_{3}, b_{4}$ form a geometric sequence with the common ratio $r=q^{10}$. Therefore,
$$
70=S_{30}=b_{1}+b_{2}+b_{3}=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right) .
$$

So $1+r+r^{2}=7, r^{2}+r-6=0$, solving this gives $r=2$ (since $r=q^{10}>0$, $r=-3$ should be discarded), hence $S_{40}=10\left(1+2+2^{2}+2^{3}\right)=150$.