If in \(A \triangle A B C\), with usual notations, \(a^2, b^2, c^2\) are in A.P. then \(\frac{\sin 3 B}{\sin B}=\)

We have \(2 b^2=a^2+c^2\)
\( \frac{\sin 3 \mathrm{~B}}{\sin \mathrm{B}}=\frac{3 \sin \mathrm{B}-4 \sin ^2 \mathrm{~B}}{\sin \mathrm{B}} \)
\( =3-4 \sin ^2 \mathrm{~B}=3-4\left(1-\cos ^2 \mathrm{~B}\right)=4 \cos ^2 \mathrm{~B}-1 \)
\( =4\left[\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ac}}\right]^2-1=4\left[\frac{\mathrm{b}^2}{2 \mathrm{ac}}\right]-1 \)
\( =\left(\frac{2 \mathrm{~b}^2}{2 \mathrm{ac}}\right)-1=\)\(\left(\frac{\mathrm{a}^2+\mathrm{c}^2}{2 \mathrm{ac}}\right)-1=\left(\frac{\mathrm{a}^2+\mathrm{c}^2}{2 \mathrm{ac}}+1\right)\left(\frac{\mathrm{a}^2+\mathrm{c}^2}{2 \mathrm{ac}}-1\right) \)
\( =\frac{(\mathrm{a}+\mathrm{c})^2(\mathrm{a}-\mathrm{c})^2}{(2 \mathrm{ac})^2} \)
\( =\left[\frac{(\mathrm{a}+\mathrm{c})(\mathrm{a}-\mathrm{c})}{2 \mathrm{ac}}\right]=\left(\frac{\mathrm{a}^2-\mathrm{c}^2}{2 \mathrm{ac}}\right)^2\)