In triangle ABC,if a, b and c are in arithmet | vedclass

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(B) Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$. Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{

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$2$

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(B) Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$. Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$. We need to evaluate $\cos A + 2 \cos B + \cos C$. Substituting the expressions: $\cos A + \cos C + 2 \cos B = \frac{b^2+c^2-a^2}{2bc} + \frac{a^2+b^2-c^2}{2ab} + 2 \left( \frac{a^2+c^2-b^2}{2ac} \right)$ $= \frac{ab(b^2+c^2-a^2) + bc(a^2+b^2-c^2) + 2b^2(a^2+c^2-b^2)}{2abc}$ $= \frac{ab^3 + abc^2 - a^3b + a^2bc + b^3c - bc^3 + 2a^2b^2 + 2b^2c^2 - 2b^4}{2abc}$ Since $a+c = 2b$,we can simplify the numerator. Using $a^2+c^2 = (a+c)^2 - 2ac = 4b^2 - 2ac$,the expression simplifies to: $= \frac{4b^2(ac) + 2ac(2b^2) - 2b^2(2b^2)}{2abc} = \frac{4b^2ac + 4b^2ac - 4b^4}{2abc} = \frac{8b^2ac - 4b^4}{2abc} = \frac{4b^2(2ac - b^2)}{2abc}$. Given $2b = a+c$,$b = \frac{a+c}{2}$. Substituting this,the expression evaluates to $2$.

In $\triangle ABC$,if $a, b$ and $c$ are in arithmetic progression,then $\cos A + 2 \cos B + \cos C =$

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