Question

If in a triangle $a{cos}^2\left(C/2\right)+c{cos}^2\left(A/2\right)=\frac{3b}{2}$, then the sides of the 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

$a{cos}^2\left(C/2\right)+c{cos}^2\left(A/2\right)=\frac{3b}{2}$
$2a{\cos }^2\frac{C}{2}+2c{\cos }^2\frac{A}{2}=3b$
$a(1+\cos C)+c(1+\cos A)=3b$
$a+a\cos C+c+c\cos A=3b$
$a+c+b=3b$ (Using Projection formula)
$\Rightarrow a+c=2b$
Hence a,b, c are in A.P.