3. For 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
3. A.

Let $b_{1}=S_{10}, b_{2}=S_{20}-S_{10}, b_{3}=S_{30}-S_{20}, b_{4}=$ $S_{\infty 0}-S_{30}$. 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 $r=q^{10}$. Therefore, $70=S_{30}=b_{1}+$
$$
\begin{array}{l}
b_{2}+b_{3}=b_{1}\left(1+r+r^{2}\right)=10\left(1+r+r^{2}\right) . \\
\quad \therefore r^{2}+r-6=0, r=2\left(\because r=q^{10}>0, \therefore r=-3\right.
\end{array}
$$

should be discarded).
Thus, $S_{40}=10\left(1+2+2^{2}+2^{3}\right)=150$.