If the angles of a $\triangle A B C$ are in $\mathrm{AP}$, then

• A

  $c^2=a^2+b^2-a b$
• B

  $a^2=b^2+c^2-a c$
• Correct answer
  C

  $b^2=a^2+c^2-a c$
• D

  $b^2=a^2+c^2$

Solution

Let the angle of $\triangle A B C$, be $(a-d), a,(a+d)$ So, $\quad a-d+a+a+d=180^{\circ} \Rightarrow a=60^{\circ}$
$\begin{aligned} \cos B & =\frac{a^2+c^2-b^2}{2 a c} \\ \Rightarrow \quad \frac{1}{2} & =\frac{a^2+c^2-b^2}{2 a c} \Rightarrow b^2=a^2+c^2-a c\end{aligned}$

Hence, the correct answer is option: 'C'.