Value of Expression Involving Side Lengths and Double Angles

Given that the angles $A$, $B$, and $C$ of a triangle are in an arithmetic progression, we have:
$2B = A + C$

Since the sum of the angles in a triangle is $180^\circ$:
$A + B + C = 180^\circ$
$3B = 180^\circ \implies B = 60^\circ$

Consequently, the sum of angles $A$ and $C$ is:
$A + C = 180^\circ - B = 120^\circ$

We are required to evaluate the expression:
$E = \frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A$

By the Sine Rule, we know that:
$\frac{a}{\sin A} = \frac{c}{\sin C} = 2R$
where $R$ is the circumradius of the triangle. Therefore, $\frac{a}{c} = \frac{\sin A}{\sin C}$ and $\frac{c}{a} = \frac{\sin C}{\sin A}$.

Substituting these ratios into the expression $E$:
$E = \frac{\sin A}{\sin C}\sin 2C + \frac{\sin C}{\sin A}\sin 2A$

Using the double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$:
$E = \frac{\sin A}{\sin C}(2\sin C\cos C) + \frac{\sin C}{\sin A}(2\sin A\cos A)$
$E = 2\sin A\cos C + 2\sin C\cos A$

Factoring out $2$ and applying the compound angle identity $\sin(A + C) = \sin A\cos C + \cos A\sin C$:
$E = 2(\sin A\cos C + \cos A\sin C)$
$E = 2\sin(A + C)$

Substituting $A + C = 120^\circ$:
$E = 2\sin 120^\circ = 2\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}$

Thus, the correct option is D.