3. Given that the three sides $a, b, c$ of $\triangle A B C$ form a geometric sequence, and the angles opposite to $a, b, c$ are $\angle A, \angle B, \angle C$ respectively. Then the range of $\sin B+\cos B$ is ( ).
(A) $\left(1,1+\frac{\sqrt{3}}{2}\right]$
(B) $\left[\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right]$
(C) $(1, \sqrt{2}]$
(D) $\left[\frac{1}{2}, \sqrt{2}\right]$
3. C.
$$
\begin{array}{l}
\text { Given } a c=b^{2}=a^{2}+c^{2}-2 a c \cos B \\
\quad \geqslant 2 a c-2 a c \cos B \\
\Rightarrow \cos B \geqslant \frac{1}{2} \Rightarrow 0<\angle B \leqslant \frac{\pi}{3} .
\end{array}
$$

Therefore, $\sin B+\cos B=\sqrt{2} \sin \left(B+\frac{\pi}{4}\right) \in(1, \sqrt{2}]$.