If in $\triangle ABC$, $\dfrac{\sin A}{\sin C} = \dfrac{\sin \left(A - B\right)}{\sin \left(B - C\right)}$, then prove that $a^2$, $b^2$ and $c^2$ are in A.P.
By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$ $\;$ where $R$ is the circumradius of $\triangle ABC$
$\implies$ $\sin A = \dfrac{a}{2 R}$, $\;$ $\sin B = \dfrac{b}{2 R}$, $\;$ $\sin C = \dfrac{c}{2 R}$
By cosine rule,
$\cos A = \dfrac{b^2 + c^2 - a^2}{2 b c}$, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
Given: $\;$ $\dfrac{\sin A}{\sin C} = \dfrac{\sin \left(A - B\right)}{\sin \left(B - C\right)}$
i.e. $\;$ $\dfrac{\sin A}{\sin C} = \dfrac{\sin A \cos B - \cos A \sin B}{\sin B \cos C - \cos B \sin C}$ $\;\;\; \cdots \; (1)$
Applying sine and cosine rule to equation $(1)$ we have,
i.e. $\;$ $\dfrac{\dfrac{a}{2 R}}{\dfrac{c}{2 R}} = \dfrac{\left(\dfrac{a}{2 R}\right) \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) - \left(\dfrac{b^2 + c^2 - a^2}{2 b c}\right) \left(\dfrac{b}{2 R}\right)}{\left(\dfrac{b}{2 R}\right) \left(\dfrac{a^2 + b^2 - c^2}{2 a b}\right) - \left(\dfrac{c^2 + a^2 - b^2}{2 c a}\right) \left(\dfrac{c}{2 R}\right)}$
i.e. $\;$ $\dfrac{a}{c} = \dfrac{\left(\dfrac{c^2 + a^2 - b^2}{c}\right) - \left(\dfrac{b^2 + c^2 - a^2}{c}\right)}{\left(\dfrac{a^2 + b^2 - c^2}{a}\right) - \left(\dfrac{c^2 + a^2 - b^2}{a}\right)}$
i.e. $\;$ $\dfrac{a}{c} = \dfrac{\dfrac{2 a^2 - 2 b^2}{c}}{\dfrac{2 b^2 - 2 c^2}{a}}$
i.e. $\;$ $\dfrac{a}{c} = \dfrac{a \left(a^2 - b^2\right)}{c \left(b^2 - c^2\right)}$
i.e. $\;$ $a^2 - b^2 = b^2 - c^2$
i.e. $\;$ $b^2 - a^2 = c^2 - b^2$
$\implies$ $a^2$, $b^2$ and $c^2$ are in A.P.
Hence proved.