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If in $\triangle \mathrm{ABC}$, with usual notations, $a \cdot \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2}$, then
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$a, b, c$ are in A.P.
$\begin{aligned} & \quad a \cdot \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2} \\ & \Rightarrow a\left(\frac{1+\cos C}{2}\right)+c\left(\frac{1+\cos A}{2}\right)=\frac{3 b}{2} \\ & \Rightarrow a+a \cos C+c+c \cos A=3 b \\ & \Rightarrow a+b+c=3 b \quad \cdots[\because b=c \cos A+a \cos C] \\ & \Rightarrow a+c=2 b \\ & \therefore \quad a, b, c \text { are in A.P. }\end{aligned}$
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