Let smallest angle $\angle A$ = $\theta$ and largest angle $\angle C$ is double of $\angle A$. $\therefore$ $\angle C$ = 2$\theta$ $\therefore$ $\angle B$ = $\pi$ - 3$\theta$ Let length of sides are a, b, c where a < b < c As a, b, c are in A.P then 2b = a + c From sin rule we can say, 2 sin B = sin A + sin C $\Rightarrow$ 2 sin ($\pi$ – 3$\theta$) = sin $\theta$ + sin 2$\theta$ $\Rightarrow$ 2 sin (3$\theta$) = sin $\theta$ + sin 2$\theta$ $\Rightarrow$ 2(3sin $\theta$ - 4sin 3 $\theta$) = sin$\theta$ + 2sin $\theta$ cos $\theta$ $\Rightarrow$ 2(3 - 4sin 2 $\theta$) = 1 + 2cos $\theta$ $\Rightarrow$ 2(3 - 4 + 4cos 2 $\theta$) = 1 + 2cos $\theta$ $\Rightarrow$ 2(4cos 2 $\theta$ - 1) = 1 + 2cos $\theta$ $\Rightarrow$ 8 cos 2 $\theta$ – 2 cos $\theta$ – 3 = 0 $\Rightarrow$ 8 cos 2 $\theta$ – 6 cos $\theta$ + 4 cos $\theta$ – 3 = 0 $\Rightarrow$ (2 cos $\theta$ + 1) ( 4 cos $\theta$ - 3) = 0 $\Rightarrow$ cos $\theta$ = ${3 \over 4}$ or cos $\theta$ = $-{1 \over 2}$ But cos $\theta$ = $-{1 \over 2}$ is not possible as $\angle C$ is acute angle. Now we have to find, a : b : c $\Rightarrow$ sin A : sin B : sin C [ from sin rule] $\Rightarrow$ sin $\theta$ : sin ($\pi$ - 3$\theta$) : sin 2$\theta$ $\Rightarrow$ sin $\theta$ : sin 3$\theta$ : sin 2$\theta$ $\Rightarrow$ sin $\theta$ : (3sin $\theta$ - 4sin 3 $\theta$) : 2sin $\theta$ cos $\theta$ $\Rightarrow$ 1 : (3 - 4sin 2 $\theta$) : 2 cos $\theta$ $\Rightarrow$ 1 : (3 - 4 + 4cos 2 $\theta$) : 2 cos $\theta$ $\Rightarrow$ 1 : (4cos 2 $\theta$ - 1) : 2 cos $\theta$ $\Rightarrow$ 1 : (4${\left( {{3 \over 4}} \right)^2}$ - 1) : 2 $\times$ ${{3 \over 4}}$ $\Rightarrow$ 1 : ${5 \over 4}$ : ${{3 \over 2}}$ $\Rightarrow$ 4 : 5 : 6