In triangle ABC the angles A, B and C are in arithmetic progression, with B the middle term, and a, $b,c$ are the sides opposite them. Then $a^2+c^2-b^2$ equals:

Being the middle term makes $B$ the average of the three angles, so $3B=180$° and $B=60$°. The cosine rule $b^2=a^2+c^2-2ac$·cosB then gives $a^2+c^2-b^2=2ac$·cos60° = $2ac\times 1/2=ac$. Forgetting the factor cos60° leaves $2ac$, halving twice gives $ac/2$, and the answer 0 would need $B=90$°, which is the Pythagorean case, not this one.