Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively, and $a{\cos }^2\left(C/2\right)+c{\cos }^2\left(A/2\right)=\frac{3b}{2}$,

then the sides of the triangle are in?

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

Answer 1

Answer: (A)

A.P.

According to Projection Rule,

$acosC+ccosA=b$

we also know that $cosA=2co{s}^2\frac{A}{2}-1,cosC=2co{s}^2\frac{C}{2}-1$

$acosC+ccosA=b$ $\Rightarrow 2aco{s}^2\frac{C}{2}-a+2cco{s}^2\frac{A}{2}-c=b\Rightarrow 2aco{s}^2\frac{C}{2}+2cco{s}^2\frac{A}{2}=a+b+c\Rightarrow aco{s}^2\frac{C}{2}+cco{s}^2\frac{A}{2}=\frac{a+b+c}{2}$

Since,

$aco{s}^2\frac{C}{2}+cco{s}^2\frac{A}{2}=\frac{3b}{2}$

we can write $\frac{a+b+c}{2}=\frac{3b}{2}$

when $a+c=2b$ which is the condition for $AP$

Therefore, Correct Answer is $A$