With usual notations in $\triangle A B C$, if $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$, then…

With usual notations in $\triangle A B C$, if $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$, then $\mathrm{a}^2, \mathrm{~b}^2, \mathrm{c}^2$ are in
  1. Not in AP
  2. HP
  3. AP
  4. GP

Solution

$\begin{aligned} & \frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)} \\ & \therefore \sin A(\sin B \cos C-\cos B \sin C)=\sin A \cos B-\cos A \sin B) \\ & \therefore \sin \mathrm{A} \sin \mathrm{B} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{B} \sin \mathrm{C}=2 \sin \mathrm{A} \cos \mathrm{B} \sin \mathrm{C} \\ & \therefore \sin \mathrm{B}(\sin \mathrm{A} \cos \mathrm{C}+\cos \mathrm{A} \sin \mathrm{C})=2 \sin \mathrm{A} \cos \mathrm{B} \sin \mathrm{C} \\ & \therefore \sin \mathrm{B}[\sin [\mathrm{A}+\mathrm{C}]=2 \sin \mathrm{A} \cos \mathrm{B} \sin \mathrm{C} \\ & \therefore \sin \mathrm{B} \sin \mathrm{B}=2 \sin \mathrm{A} \cos \mathrm{B} \sin \mathrm{C} \quad \ldots[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi] \\ & \frac{1}{2 \cos B}=\frac{\sin A}{\sin B} \times \frac{\sin C}{\sin B} \\ & \text { By sine rule, we know that } \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ & \therefore \frac{1}{2\left(\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ac}}\right)}=\frac{\mathrm{a}}{\mathrm{b}} \times \frac{\mathrm{a}}{\mathrm{c}} \Rightarrow \frac{\mathrm{ac}}{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}=\frac{\mathrm{ac}}{\mathrm{b}^2} \\ & \therefore \mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2=\mathrm{b}^2 \Rightarrow 2 \mathrm{~b}^2=\mathrm{a}^2+\mathrm{c}^2 \\ & \end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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