If the angles of a triangle \[ABC\] are in arithmetic progression, then which of the following statement is true?
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. \[{b^2} = {a^2} + {c^2}\]

Hint: Here, the angles of the given triangle are in the arithmetic progression. We will use the rule of the arithmetic progression for the 3 numbers in an AP to calculate the value of the middle angle. Then, we will apply the law of cosines to that angle. In the end, solve the equation to get the required answer.

Formula used:
If the three numbers \[a, b\] and \[c\] are in the arithmetic progression, then \[ b = \dfrac{{a + c}}{2}\].
Law of cosines: \[\cos B = \dfrac{{{c^2} + a^{2} - {b^2}}}{{2ac}}\].

Complete step by step solution:
Given:
The angles of the triangle \[ABC\] are in the arithmetic progression.

We know that the sum of the internal angles of any triangle is \[180^ {\circ }\].
So, \[A + B + C = 180^ {\circ }\] \[.....\left( 1 \right)\]

Since the angles of the triangle are in the arithmetic progression.
We get,
\[B = \dfrac{{A + C}}{2}\]
Substitute the value of \[A + C\] by using the equation \[\left( 1 \right)\].
\[B = \dfrac{{180^ {\circ } - B}}{2}\]
\[ \Rightarrow 3B = 180^ {\circ }\]
\[ \Rightarrow B = 60^ {\circ }\]

Now apply the laws of cosines to angle B.
We get,
\[\cos B = \dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}\]
Substitute the value of the angle in the above equation.
\[\cos 60^ {\circ } = \dfrac{{a{}^2 + {c^2} - {b^2}}}{{2ac}}\]
\[ \Rightarrow \dfrac{1}{2} = \dfrac{{a{}^2 + {c^2} - {b^2}}}{{2ac}}\]
\[ \Rightarrow ac = a{}^2 + {c^2} - {b^2}\]
\[ \Rightarrow {b^2} = a{}^2 + {c^2} - ac\]
Hence the correct option is B.

Note: While solving the question keep in mind that if the numbers are in the arithmetic progression, then the difference between each consecutive term is a constant value. And if the three terms are in the arithmetic progression, then the sum of the first and last terms is twice the middle term.