3. Given $A_{1}, A_{2}, \cdots ; B_{1}, B_{2}, \cdots$ are geometric sequences. Can $A_{1}+B_{1}, A_{2}+B_{2}, A_{3}+B_{3}$ and $A_{1}+B_{4}$ determine $A_{5}+B_{5}$?
3. Let $A_{1}=a p^{i-1}, B_{1}=b q^{i-1}(1-i-5)$,
$S_{1}=A_{1}+B_{1}$, then
$$
S_{i+1}(p+q)=S_{i+2}+S_{i} p q(1 \leqslant i \leqslant 3) .
$$

If $S_{1} S_{3} \neq S_{2}^{2}$, then from
$$
\left\{\begin{array}{l}
S_{2}(p+q)-S_{1} p q=S_{3}, \\
S_{3}(p+q)-S_{2} p q=S_{4},
\end{array}\right.
$$

$p+q$ and $p q$ can be uniquely determined, thus
$S_{5}=S_{4}(p+q)-S_{3} p q$ is uniquely determined.
If $S_{1} S_{3}=S_{2}^{2}$, i.e., $a b(p-q)^{2}=0$.

If $a=0, b=0$ or $p=q$, then $S_{1}=b q^{1-1}, a p^{1-1}$ or $(a+b) p^{i-1}$. In this case, if $S_{1}=0$, then $S_{5}=0$; if $S_{1} \neq 0$, then $S_{5}=\frac{S_{3}^{2}}{S_{1}}$.
In any case, $S_{5}$ can be uniquely determined from $S_{1}$ to $S_{4}$.