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In a $\triangle A B C$, if $a \cos ^2 \frac{C}{2}+c \cos ^2 \frac{A}{2}=\frac{3 b}{2}$, then the sides of the triangle are in
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The correct answer is:
an arithmetic progression
In $\triangle A B C, a \cos ^2 \frac{c}{2}+a \cos ^2 \frac{A}{2}=\frac{3 b}{2}$
$\begin{array}{lc}
\Rightarrow & a\left\{\sqrt{\frac{s(s-c)}{a b}}\right\}^2+c\left\{\frac{\sqrt{s(s-a)}}{b c}\right\}=\frac{3 b}{2} \\
\Rightarrow & a \cdot \frac{s(s-c)}{a b}+c \cdot \frac{s(s-a)}{b c}=\frac{3 b}{2} \\
\Rightarrow & \frac{s(s-c)}{b}+\frac{s(s-a)}{b}=\frac{3 b}{2} \\
\Rightarrow & \frac{s}{b}(s-c+s-a)=\frac{3 b}{2} \\
\Rightarrow & 2 s(2 s-a-c)=3 b^2 \\
\Rightarrow & 2 s(a+b+c-a-c)=3 b^2 \\
\Rightarrow & \quad[\because 2 s=a+b+c] \\
\Rightarrow & 2 s \cdot b=3 b^2 \\
\Rightarrow & 2 s=3 b \\
\Rightarrow & a+b+c=3 b \\
\Rightarrow & 2 b=a+c
\end{array}$
$\Rightarrow a, b, c$ are in arithmetic progression.
$\begin{array}{lc}
\Rightarrow & a\left\{\sqrt{\frac{s(s-c)}{a b}}\right\}^2+c\left\{\frac{\sqrt{s(s-a)}}{b c}\right\}=\frac{3 b}{2} \\
\Rightarrow & a \cdot \frac{s(s-c)}{a b}+c \cdot \frac{s(s-a)}{b c}=\frac{3 b}{2} \\
\Rightarrow & \frac{s(s-c)}{b}+\frac{s(s-a)}{b}=\frac{3 b}{2} \\
\Rightarrow & \frac{s}{b}(s-c+s-a)=\frac{3 b}{2} \\
\Rightarrow & 2 s(2 s-a-c)=3 b^2 \\
\Rightarrow & 2 s(a+b+c-a-c)=3 b^2 \\
\Rightarrow & \quad[\because 2 s=a+b+c] \\
\Rightarrow & 2 s \cdot b=3 b^2 \\
\Rightarrow & 2 s=3 b \\
\Rightarrow & a+b+c=3 b \\
\Rightarrow & 2 b=a+c
\end{array}$
$\Rightarrow a, b, c$ are in arithmetic progression.
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