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(A) The angles of $	riangle ABC$ are in arithmetic progression $(AP)$.Since $A+B+C=180^{\circ}$ and $2B=A+C$,we have $3B=180^{\circ}$,which implies $

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$b^2=a^2+c^2-ac$

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(A) The angles of $	riangle ABC$ are in arithmetic progression $(AP)$.Since $A+B+C=180^{\circ}$ and $2B=A+C$,we have $3B=180^{\circ}$,which implies $B=60^{\circ}$.Using the Law of Cosines,$\cos B = \frac{a^2+c^2-b^2}{2ac}$.Substituting $B=60^{\circ}$,we get $\cos 60^{\circ} = \frac{a^2+c^2-b^2}{2ac}$.Since $\cos 60^{\circ} = \frac{1}{2}$,we have $\frac{1}{2} = \frac{a^2+c^2-b^2}{2ac}$.Multiplying both sides by $2ac$,we get $ac = a^2+c^2-b^2$.Therefore,$b^2 = a^2+c^2-ac$.

If the angles $A, B$ and $C$ of $\triangle ABC$ are in arithmetic progression,then

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