If  $\frac{\sin 3B}{\sin B}={\left(\frac{a^2-c^2}{2ac}\right)}^2$, then  $a^2,b^2,c^2$ are in

$\frac{\sin 3B}{\sin B}={\left(\frac{a^2-c^2}{2ac}\right)}^2$

$\frac{3\sin B-4{\sin }^{3}B}{\sin B}={\left(\frac{a^2-c^2}{2ac}\right)}^2$

$3-4{\sin }^{2}B={\left(\frac{a^2-c^2}{2ac}\right)}^2$

$3-4(1-{\cos }^{2}B)={\left(\frac{a^2-c^2}{2ac}\right)}^2$

$-1+4{\left(\frac{a^2+c^2-b^2}{2ac}\right)}^2=\frac{{(a^2-c^2)}^2}{4a^2{c}^2}$

$\Rightarrow -4a^2{c}^2+4{(a^2+c^2-b^2)}^2={(a^2-c^2)}^2$

$4{(a^2+c^2-b^2)}^2={(a^2-c^2)}^2+4a^2{c}^2$

$4{(a^2+c^2-b^2)}^2={(a^2+c^2)}^2$

$2(a^2+c^2-b^2)=a^2+c^2$

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

$\therefore a^2,b^2,c^2$  are in A.P.