(2) In $\triangle A B C$, the three interior angles $A, B, C$ form an arithmetic sequence, and $\sin A, \sin B, \sin C$ form a geometric sequence, then $\triangle A B C($.
(A) is a right triangle, but not an isosceles triangle
(B) is neither an isosceles triangle nor a right triangle
(C) is an isosceles right triangle
(D) is an equilateral triangle
2. Since $B=60^{\circ}, \sin ^{2} B=\sin A \sin C$, we have
$$
\sin ^{2} 60^{\circ}=\sin A \sin \left(120^{\circ}-A\right),
$$

Therefore,
$$
\begin{aligned}
\frac{3}{4} & =\sin A\left(\frac{\sqrt{3}}{2} \cos A+\frac{1}{2} \sin A\right) \\
& =\frac{\sqrt{3}}{2} \sin A \cos A+\frac{1}{2} \sin ^{2} A \\
& =\frac{\sqrt{3}}{4} \sin 2 A+\frac{1-\cos 2 A}{4}
\end{aligned}
$$

Thus $\sin \left(2 A-30^{\circ}\right)=1$, and since $0^{\circ}<A<120^{\circ}$, it is easy to see that $2 A-30^{\circ}=90^{\circ}$, which means $A=$ $60^{\circ}$. Therefore, $\triangle A B C$ is an equilateral triangle. Hence, the answer is D.