Question

In a $△ABC,$ if $\cos A+2\cos B+\cos C=2,$ then the sides of triangle are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P.$
• D. None of these

Answer 1

The correct option is A $A.P.$

Given,
$\cos A+2\cos B+\cos C=2\Rightarrow \cos A+\cos C=2(1-\cos B)\Rightarrow 2\cos \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)=4{\sin }^2\frac{B}{2}[As,\cos \left(\frac{A+C}{2}\right)=\cos (\frac{\pi }{2}-\frac{B}{2})=\sin \frac{B}{2}]\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\cos \left(\frac{A+C}{2}\right)\Rightarrow \cos \frac{A}{2}\cdot \cos \frac{C}{2}+\sin \frac{A}{2}\cdot \sin \frac{C}{2}=2\cos \frac{A}{2}\cdot \cos \frac{C}{2}-2\sin \frac{A}{2}\cdot \sin \frac{C}{2}\Rightarrow \cot \frac{A}{2}\cdot \cot \frac{C}{2}=3\Rightarrow \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}\cdot \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=3\Rightarrow \frac{s}{(s-b)}=3\Rightarrow s=3s-3b\Rightarrow 2s=3b\Rightarrow a+c=2b$
$\therefore a,b,c$ are in $A.P.$