Given, in a triangle $A B C$ if the angles $A, B, C$ are in $A P$.
$\Rightarrow 2 B=A+C$
We know that, the sum of the angles of the triangle is $180^{\circ}$.
Given, in a triangle $A B C$ if the angles $A, B, C$ are in $A P$.
$\Rightarrow A+B+C=180^{\circ}$
$\Rightarrow 2 B+B=180^{\circ}$
$\Rightarrow 3 B=180^{\circ}$
$\Rightarrow B=60^{\circ}$
By cosine formula, we have
$\cos (B)=\frac{a^2+c^2-b^2}{2 a c}$
$\cos (B)=\frac{a^2+c^2-b^2}{2 a c}$
$\Rightarrow \cos \left(60^{\circ}\right)=\frac{a^2+c^2-b^2}{2 a c}$
$\Rightarrow \frac{1}{2}=\frac{\mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2}{2 \mathrm{ac}}$
$\Rightarrow \mathrm{a}^2+\mathrm{c}^2-\mathrm{b}^2=\mathrm{ac}$
$\Rightarrow b^2=a^2+c^2-a c$
Hence, In a triangle $\mathrm{ABC}$ if the angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in $\mathrm{AP}, \mathrm{b}^2=\mathrm{a}^2+\mathrm{c}^2-\mathrm{ac}$ is the correct statement.