1. Given an arithmetic sequence $\left\{a_{n}\right\}$ with common difference $d \neq 0$, and a geometric sequence $\left\{b_{n}\right\}$ with common ratio $q$ being a positive rational number less than 1. If $a_{1}=d, b_{1}=d^{2}$, and $\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{b_{1}+b_{2}+b_{3}}$ is a positive integer, then $q=$ $\qquad$
$-1 . \frac{1}{2}$.
From the problem, we know
$$
\begin{array}{l}
\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}{b_{1}+b_{2}+b_{3}} \\
=\frac{a_{1}^{2}+\left(a_{1}+d\right)^{2}+\left(a_{1}+2 d\right)^{2}}{b_{1}+b_{1} q+b_{1} q^{2}} \\
=\frac{14}{1+q+q^{2}}=m\left(m \in \mathbf{Z}_{+}\right) .
\end{array}
$$

Then $1+q+q^{2}=\frac{14}{m}$
$$
\begin{aligned}
\Rightarrow & q=-\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{14}{m}-1} \\
& =-\frac{1}{2}+\sqrt{\frac{56-3 m}{4 m}} .
\end{aligned}
$$

Since $q$ is a positive rational number less than 1, we have
$$
1<\frac{14}{m}<3 \Rightarrow 5 \leqslant m \leqslant 13 \text {. }
$$

Also, $\frac{56-3 m}{4 m}$ is the square of some rational number, thus
$$
q=\frac{1}{2} .
$$