Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $c{\cos }^2\frac{A}{2}+a{\cos }^2\frac{C}{2}=\frac{3b}{2},$ then $a,b,c$ are in?

• A. $GP$
• B. $HP$
• C. $AP$
• D. none of these

Answer 1

Answer: (C)

$AP$

$a{\cos }^2\left(C/2\right)+c{\cos }^2\left(A/2\right)=\frac{3b}{2}$

$\Rightarrow a2{\cos }^2\left(C/2\right)+c2{\cos }^2\left(A/2\right)=3b$

$\Rightarrow a(1+\cos C)+c(1+\cos A)=3b$

$\Rightarrow a(1+\frac{a^2+b^2-c^2}{2ab})+c(1+\frac{b^2+c^2-a^2}{2bc})=3b$

$\Rightarrow (a+b{)}^2-c^2+(b+c{)}^2-a^2=6b^2$

$\Rightarrow (a+b+c)(a+b-c+b+c-a)=6b^2$

$\Rightarrow (a+b+c)\left(2b\right)=6b^2$

$\Rightarrow a+c=2b\therefore a,b,c\in$A.P