The angles of a triangle are in AP.

Given that, the angles of a triangle are in AR

Let A, B and C are angles of a Δ ABC.

$\therefore$ $B=\frac{A+C}{2}$

$\Rightarrow \quad 2 B=A+C$ ...(i)

We know that, sum of all interior angles of a $\triangle A B C=180^{\circ}$

$A+B+C=180^{\circ}$

$\Rightarrow \quad 2 B+B=180^{\circ}$ [from Eq. (i)]

$\Rightarrow$ $3 B=180^{\circ} \Rightarrow B=60^{\circ}$

Let the greatest and least angles are $A$ and $C$ respectively. 

$A=2 C$ [by condition] ... (ii)

Now, put the values of $B$ and $A$ in Eq. (i), we get

$2 \times 60=2 C+C$

$\Rightarrow$ $120=3 C \Rightarrow C=40^{\circ}$

Put the value of $C$ in Eq. (ii), we get

$A=2 \times 40^{\circ} \Rightarrow A=80^{\circ}$

Hence, the required angles of triangle are $80^{\circ}, 60^{\circ}$ and $40^{\circ}$.