3. For the 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
3. 【Analysis and Solution】Let $b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, b_{3}=S_{30}-S_{20}, b_{4}=S_{40}-S_{30}$, and 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 a common ratio of $r=q^{10}$. Therefore, $70=S_{10}=b_{1}+b_{2}+b_{3}=b_{1}\left(1+r+r^{2}\right)=$ $10\left(1+r+r^{2}\right)$.
$\therefore r^{2}+r-6=0, r=2\left(\because r=q^{10}>0, \therefore r=-3\right.$ should be discarded $)$.
Thus, $S_{40}=10\left(1+2+2^{2}+2^{3}\right)=150$.