If in a triangle ABC,b cos^2 frac{A}{2} + a c | vedclass

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(D) Given the equation: $b \cos^2 \frac{A}{2} + a \cos^2 \frac{B}{2} = \frac{3}{2} c$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,

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(D) Given the equation: $b \cos^2 \frac{A}{2} + a \cos^2 \frac{B}{2} = \frac{3}{2} c$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:$\frac{b}{2}(1 + \cos A) + \frac{a}{2}(1 + \cos B) = \frac{3c}{2}$.Multiplying by $2$:$b + b \cos A + a + a \cos B = 3c$.Since $a = b \cos C + c \cos B$ and $b = a \cos C + c \cos A$,we have $b \cos A + a \cos B = c$.Substituting this into the equation:$(a + b) + c = 3c$.$a + b = 2c$.This implies that $a, c, b$ are in $A.P.$ However,the question asks for the relationship between $a, b, c$. Given $a+b=2c$,this is the condition for $a, c, b$ to be in $A.P.$ If the question implies the sequence $a, b, c$,none of the standard progressions apply directly unless specified.

If in a triangle $ABC$,$b \cos^2 \frac{A}{2} + a \cos^2 \frac{B}{2} = \frac{3}{2} c$,then $a, b, c$ are in:

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