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(D) Let the angles of $	riangle ABC$ be $A-d, A, A+d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.Thus,t

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$\frac{a \pm \sqrt{4b^2-3a^2}}{2}$

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(D) Let the angles of $	riangle ABC$ be $A-d, A, A+d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.Thus,the angles are $60^{\circ}-d, 60^{\circ}, 60^{\circ}+d$.Using the Law of Cosines for the angle $60^{\circ}$:$\cos 60^{\circ} = \frac{a^2+c^2-b^2}{2ac} = \frac{1}{2}$$a^2+c^2-b^2 = ac$$c^2 - ac + (a^2-b^2) = 0$Solving this quadratic equation for $c$ using the quadratic formula:$c = \frac{a \pm \sqrt{a^2 - 4(a^2-b^2)}}{2}$$c = \frac{a \pm \sqrt{4b^2-3a^2}}{2}$

The angles of $\triangle ABC$ are in an arithmetic progression. If the larger sides $a, b$ satisfy the relation $\frac{\sqrt{3}}{2} < \frac{b}{a} < 1$,then the possible values of the smallest side $c$ are

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