If the sides of a triangle $a, b, c$ are in A.P., then with usual notations, a $\cos ^2…
If the sides of a triangle $a, b, c$ are in A.P., then with usual notations, a $\cos ^2 \frac{\mathrm{C}}{2}+\mathrm{c} \cos ^2 \frac{\mathrm{~A}}{2}$ is
$\frac{3 a}{2}$
$\frac{3 \mathrm{c}}{2}$
$\frac{3 b}{2}$
$\frac{\mathrm{a}+\mathrm{c}}{2}$
Solution
Since $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A. P.,
$\begin{aligned}
& 2 b=a+c \\
& a \cos ^2\left(\frac{C}{2}\right)+c \cos ^2\left(\frac{A}{2}\right) \\
& =\frac{a(1+\cos C)}{2}+\frac{c(1+\cos A)}{2} \\
& =\frac{a+c+a \cos C+c \cos A}{2} \\
& =\frac{a+c+b}{2} \\
& =\frac{2 b+b}{2}=\frac{3 b}{2}
\end{aligned}$