5. As shown in the figure, it is a $5 \times$ number table,
$$
\begin{array}{lllll}
a_{11} & a_{12} & a_{13} & a_{14} & a_{15} \\
a_{21} & a_{22} & a_{23} & a_{24} & a_{25} \\
a_{31} & a_{32} & a_{33} & a_{34} & a_{35} \\
a_{41} & a_{42} & a_{43} & a_{44} & a_{45} \\
a_{51} & a_{52} & a_{53} & a_{54} & a_{55}
\end{array}
$$

where $a_{i 1}, a_{i 2}, a_{i 3}, a_{i 4}, a_{i 5}(1 \leq i \leq 5)$ form an arithmetic sequence, and $a_{1 j}, a_{2 j}, a_{3 j}, a_{4 j}, a_{5 j}(1 \leq j \leq 5)$ form a geometric sequence, with the common ratio being the same for each column. Given that $a_{24}=4, a_{41}=-2, a_{43}=10$, then $a_{11} \times a_{55}=$
Solution: $d=\frac{a_{44}-a_{42}}{2}=6$, thus $a_{42}=a_{41}+d=4, a_{44}=a_{43}+d=16, a_{45}=a_{44}+d=22$,
hence $q^{2}=\frac{a_{44}}{a_{24}}=4$, so $q= \pm 2$,
we get $a_{11}=\frac{a_{41}}{q^{3}}=\mp \frac{1}{2}, a_{55}=a_{45} \cdot q= \pm 22$,
therefore $a_{11} \times a_{55}=-11$.