5. Given three distinct integers $x, y, z$ whose sum lies between 40 and 44. If $x, y, z$ form an arithmetic sequence with a common difference of $d$, and $x+y, y+z, z+x$ form a geometric sequence with a common ratio of $q$, then $d q=$ $\qquad$
5. 42 .
$$
\begin{array}{l}
\text { Given } x=y-d, z=y+d \\
\Rightarrow x+y=2 y-d, y+z=2 y+d \\
\Rightarrow z+x=2 y . \\
\text { Also, }(x+y)(z+x)=(y+z)^{2} \\
\Rightarrow 2 y(2 y-d)=(2 y+d)^{2} \\
\Rightarrow d(d+6 y)=0 .
\end{array}
$$

Since $d \neq 0$, we have $d=-6 y$.
$$
\begin{array}{l}
\text { Also, } 40<x+y+z=3 y<44 \\
\Rightarrow \frac{40}{3}<y<\frac{44}{3} \Rightarrow y=14 \\
\Rightarrow d=-6 y=-84 \\
\Rightarrow q=\frac{y+z}{x+y}=\frac{2 y+d}{2 y-d}=-\frac{1}{2} \\
\Rightarrow d p=42 .
\end{array}
$$