A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
\[\frac{\sin A}{\sin C}=\frac{\sin A\cos B-\cos A\sin B}{\sin B\cos C-\cos B\sin C}\] Þ \[\frac{a}{c}=\frac{a\cos B-b\cos A}{b\cos C-c\cos B}\], (Using sine formula) Þ \[\,ab\cos C-ac\cos B=ac\cos B-bc\cos A\] Þ \[\,\,ab\cos C+bc\cos A=2ac\cos B\] Þ \[\,\,\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2}+\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2}=\frac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{1}\] Þ \[{{b}^{2}}={{c}^{2}}+{{a}^{2}}-{{b}^{2}}\]Þ \[{{b}^{2}}=\frac{{{c}^{2}}+{{a}^{2}}}{2}\] \[\Rightarrow {{a}^{2}},\,{{b}^{2}},\,{{c}^{2}}\]are in A.P.
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