A) \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}-ab\]
B) \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\]
C) \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}-ac\]
D) \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}\]
Correct Answer: B
Solution :
A, B, C are in A. P. then angle\[B={{60}^{o}},\] \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\], \[\left\{ \begin{align} & \text{since }A+B+C={{180}^{o}}\,\,\text{and} \\ & \text{ }A+C=2B\Rightarrow B={{60}^{o}} \\ \end{align} \right\}\] Þ \[\frac{1}{2}=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\Rightarrow {{a}^{2}}+{{c}^{2}}-{{b}^{2}}=ac\] Þ \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\].
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