Question

If $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)},$ then $a^2,b^2,c^2$ are in

• A. A.P
• B. G.P
• C. H.P
• D. none of these

Answer 1

The correct option is A A.P

$\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$
$\Rightarrow \frac{\sin \pi -(B+C)}{\sin \pi -(A+B)}=\frac{\sin (A-B)}{\sin (B-C)}$
$\Rightarrow \frac{\sin (B+C)}{\sin (A+B)}=\frac{\sin (A-B)}{\sin (B-C)}$
$\Rightarrow \sin (B+C)(B-C)=\sin (A+B)\sin (A-B)$
$\Rightarrow {\sin }^{2}B-{\sin }^{2}C={\sin }^{2}A-{\sin }^{2}B$
$\Rightarrow \frac{b^2}{k^2}-\frac{c^2}{k^2}=\frac{a^2}{k^2}-\frac{b^2}{k^2}$
$\Rightarrow 2b^2=a^2+c^2$
Hence $a^2,b^2,c^2$ are in A.P