1. In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If the sizes of angles $A, B, C$ form a geometric sequence, and $b^{2}-a^{2}=a c$, then the radian measure of angle $B$ is $\qquad$.
1. $\because A, B, C$ form a geometric sequence,
$\therefore$ without loss of generality,
$$
A=\frac{1}{q} B, C=q B .
$$

From $A+B+C=\pi$ we have
$$
B=\frac{q \pi}{q^{2}+q+1} .
$$

Given $b^{2}-a^{2}=a c$.
By the cosine rule,
$$
b^{2}=a^{2}+c^{2}-2 a c \cos B,
$$

we get
$$
\begin{aligned}
\because \quad c \neq 0, \quad a c & =c^{2}-2 a c \cos B . \\
\quad a & =c-2 a \cos B .
\end{aligned}
$$

Using the sine rule, we obtain
$$
\sin A=\sin C-2 \sin A \cos B,
$$

thus
$$
\begin{array}{l}
\sin A=\sin C-(\sin (A+B)+\sin (A-B)] . \\
\therefore \quad \sin A=\sin (B-A) .
\end{array}
$$

Solving this, we get
$$
\begin{array}{l}
A=\frac{1}{2} B, \\
\therefore \quad q=2, \quad B=\frac{2 \pi}{7} .
\end{array}
$$