Question

In $△ABC$, if $\cos A+2\cos B+\cos C=2$ then $a,b,c$ are in

• A. $A.P$
• B. $H.P$
• C. $G.P$
• D. $A.G.P$

Answer 1

The correct option is A $A.P$

$formulaused$
$\cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-C}{2}\right)$
$2\sin A\cos A=\sin 2A$
$Given$
$\mathrm{Cos}A+2\mathrm{Cos}B+\mathrm{Cos}C=2$
$\Rightarrow \mathrm{Cos}A+\mathrm{Cos}C=2-2\mathrm{Cos}B$
$\Rightarrow 2\mathrm{Cos}\left(\frac{A+C}{2}\right)\mathrm{Cos}\left(\frac{A-C}{2}\right)=2(1-\mathrm{Cos}B)$
$\Rightarrow 2\mathrm{Cos}\left(\frac{\pi -B}{2}\right)\mathrm{Cos}\left(\frac{A-C}{2}\right)=2\left(2{\sin }^2\frac{B}{2}\right)$
$\Rightarrow \sin \left(\frac{B}{2}\right)\cos \left(\frac{A-C}{2}\right)=2{\sin }^2\left(\frac{B}{2}\right)$
$\Rightarrow \cos \left(\frac{A-C}{2}\right)=2\sin \left(\frac{B}{2}\right)$
$\Rightarrow \cos \left(\frac{B}{2}\right)\cos \left(\frac{A-C}{2}\right)=2\sin \left(\frac{B}{2}\right)\cos \left(\frac{B}{2}\right)$
$\Rightarrow \cos \left(\frac{\pi -(A+C)}{2}\right)\cos \left(\frac{A-C}{2}\right)=\sin B$
$\Rightarrow \sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=\sin B$
$\Rightarrow 2\sin \left(\frac{A+C}{2}\right)\cos \left(\frac{A-C}{2}\right)=2\sin B$
$\Rightarrow \sin A+\sin C=2\sin B.........\left(i\right)$
$nowfrom\sin erule$
$\frac{\mathrm{Sin}A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k$
$\Rightarrow \sin A=ka,\sin B=kb,\sin C=kc$
$so\left(i\right)becomes$
$\Rightarrow ka+kc=2kb$
$\Rightarrow \frac{a+c}{2}=b$
$HencesidesofthetriangleareinA.P$