A) A. P.
B) G. P.
C) H. P.
D) None of these
Correct Answer: A
Solution :
\[\cos A+\cos C=4{{\sin }^{2}}\frac{1}{2}B\] Þ \[2\cos \frac{A+C}{2}\cos \frac{A-C}{2}=4{{\sin }^{2}}\frac{B}{2}\] Þ \[\cos \frac{A+C}{2}\cos \frac{A-C}{2}=2{{\sin }^{2}}\frac{B}{2}\] Þ \[\cos \left( \frac{A-C}{2} \right)=2\sin \frac{B}{2}\] Þ \[\cos \frac{A}{2}\cos \frac{C}{2}+\sin \frac{A}{2}\sin \frac{C}{2}=2\sin \frac{B}{2}\] Þ \[\sqrt{\frac{s(s-a)}{bc}}\sqrt{\frac{s(s-c)}{ab}}+\sqrt{\frac{(s-b)(s-c)}{bc}}\sqrt{\frac{(s-a)(s-b)}{ab}}\] \[=2\sqrt{\frac{(s-a)(s-c)}{ac}}\] \[\frac{\sqrt{(s-a)\,(s-c)}}{ac}+\frac{s-b}{b}\sqrt{\frac{(s-c)\,\text{ (}s-a\text{)}}{ac}}\]\[=2\sqrt{\frac{(s-a)\,\text{(}s-c)}{ac}}\] Þ \[\frac{s}{b}+\frac{s-b}{b}=2\] Þ \[a+c=2b\]Þ \[a,b,c\] are in A. P.
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