If the sides of a triangle are in AP, and the greatest angle of the triangle is double the smallest angle, the ratio of the sides of the triangle is

Let the sides of the triangle be a-d, a and a+d, with a > d>0. 

Clearly, a -d is the smallest side and a + d is the largest side. 

So, A is the smallest angle and C is the largest angle. It is given that C = 2A. 

Thus, the angles of the triangle are A, 2A and  $\mathrm{\pi }$-3A. 

Applying the law of sines, we obtain 

$\begin{array}{rl} & \frac{\mathrm{a}-\mathrm{d}}{\sin \mathrm{A}}=\frac{\mathrm{a}}{\sin (\mathrm{\pi }-3\mathrm{A})}=\frac{\mathrm{a}+\mathrm{d}}{\sin 2\mathrm{A}}\\ \Rightarrow & \frac{\mathrm{a}-\mathrm{d}}{\sin \mathrm{A}}=\frac{\mathrm{a}}{\sin 3\mathrm{A}}=\frac{\mathrm{a}+\mathrm{d}}{\sin 2\mathrm{A}}\\ \Rightarrow & \frac{\mathrm{a}-\mathrm{d}}{\sin \mathrm{A}}=\frac{\mathrm{a}}{3\sin \mathrm{A}-4{\sin }^3\mathrm{A}}=\frac{\mathrm{a}+\mathrm{d}}{2\sin \mathrm{Acos}\mathrm{A}}\\ \Rightarrow & \frac{\mathrm{a}-\mathrm{d}}{1}=\frac{\mathrm{a}}{3-4{\sin }^2\mathrm{A}}=\frac{\mathrm{a}+\mathrm{d}}{2\cos \mathrm{A}}\\ \Rightarrow & 3-4{\sin }^2\mathrm{A}=\frac{\mathrm{a}}{\mathrm{a}-\mathrm{d}}\text{and}2\cos \mathrm{A}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a}-\mathrm{d}}\\ \Rightarrow & 4{\cos }^2\mathrm{A}-1=\frac{\mathrm{a}}{\mathrm{a}-\mathrm{d}}\text{and}2\cos \mathrm{A}=\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a}-\mathrm{d}}\end{array}$

$\Rightarrow {\left(\frac{\mathrm{a}+\mathrm{d}}{\mathrm{a}-\mathrm{d}}\right)}^2-1=\frac{\mathrm{a}}{\mathrm{a}-\mathrm{d}}\Rightarrow \mathrm{a}=5\mathrm{d}$

Thus, the sides of the triangle are a-d, a, a + d    i.e. 4d, 5d, 6d.

Hence, the ratio of the sides of the triangle is 4d :5d: 6d     i.e. 4:5:6.