4. Given that the three interior angles $A, B, C$ of $\triangle A B C$ form an arithmetic sequence, and the corresponding sides are $a, b, c$, and $a, c, \frac{4}{\sqrt{3}} b$ form a geometric sequence, then $S_{\triangle A B C}: a^{2}=$ $\qquad$ .
The answer is $\frac{\sqrt{3}}{2}$.
Analysis: Since $A, B, C$ form an arithmetic sequence, $2B = A + C, 3B = A + B + C = 180^{\circ}$, thus $B = 60^{\circ}$. Also, since $a, c, \frac{4}{\sqrt{3}} b$ form a geometric sequence, we have $c = q a, b = \frac{\sqrt{3} q^{2} a}{4}$. By the Law of Sines,
$$
\frac{a}{\sin A} = \frac{\sqrt{3} q^{2} a}{4 \sin 60^{\circ}} = \frac{q a}{\sin \left(120^{\circ} - A\right)}
$$

Rearranging gives
$$
\sin A = \frac{2}{q^{2}}, \quad \frac{\sqrt{3}}{2} \cos A = \frac{2}{q} - \frac{1}{q^{2}}, \quad (q - 2)\left[3 q^{3} + 5 q^{2} + 4 + (q - 2)^{2}\right] = 0
$$

Therefore, $q = 2, \sin A = \frac{1}{2}, A = 30^{\circ}, C = 90^{\circ}$, hence $S_{\triangle A B C} = \frac{1}{2} a b = \frac{\sqrt{3}}{2} a^{2}$.