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,
$\therefore$ we can assume
$$
A=\frac{1}{q} B, \quad C=q B .
$$

From $A+B+C=\pi$ we know

Given
$$
\begin{array}{c}
B=\frac{q \pi}{q^{2}+q+1} . \\
b^{2}-a^{2}=a c .
\end{array}
$$

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

we get
$$
a c=c^{2}-2 a c \cos B \text {. }
$$
$$
\begin{array}{lc}
\because & c \neq 0, \\
\therefore & a=c-2 a \cos B .
\end{array}
$$

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

thus $\sin A=\sin C-[\sin (A+B)+\sin (A-B)]$.
$$
\therefore \quad \sin A=\sin (B-A) .
$$

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