If the lengths of the sides of a triangle are in A.P and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is:

Let the three sides, taken in ascending order, be $$(b-d),\;b,\;b+d$$ where $$b>d>0$$ so that they form an arithmetic progression of common difference $$d$$. Call the angles opposite these sides $$\alpha,\;\beta,\;\gamma$$ respectively, so $$\gamma$$ is the largest and $$\alpha$$ is the smallest.

The statement “the greatest angle is double the smallest” translates to $$\gamma = 2\alpha.$$ Because a triangle’s angles sum to $$180^\circ,$$ we have

$$\alpha+\beta+\gamma=\alpha+\beta+2\alpha=180^\circ\;\Longrightarrow\;\beta = 180^\circ-3\alpha.$$

Invoke the Law of Sines:

$$\frac{b-d}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{b+d}{\sin2\alpha}.$$

For convenience set the smallest side $$b-d$$ equal to one unit, i.e. $$b-d=1.$$ Then, writing the equalities in pairwise form,

$$\frac{b}{1}= \frac{\sin\beta}{\sin\alpha}$$ $$\text{and}$$ $$\frac{b+d}{1}= \frac{\sin2\alpha}{\sin\alpha}.$$

First simplify the right-hand ratios. Using $$\beta=180^\circ-3\alpha$$ and the identity $$\sin(180^\circ-\theta)=\sin\theta,$$

$$\frac{\sin\beta}{\sin\alpha}=\frac{\sin3\alpha}{\sin\alpha}.$$ Write $$\sin3\alpha=3\sin\alpha-4\sin^3\alpha,$$ so

$$\frac{\sin3\alpha}{\sin\alpha}=3-4\sin^2\alpha.$$

Similarly, $$\frac{\sin2\alpha}{\sin\alpha}= \frac{2\sin\alpha\cos\alpha}{\sin\alpha}=2\cos\alpha.$$

Hence

$$\boxed{b = 3-4\sin^2\alpha}$$ and $$\boxed{\,b+d = 2\cos\alpha\,}.$$

Because $$\sin^2\alpha + \cos^2\alpha =1,$$ convert the first boxed equation entirely into $$\cos\alpha$$ :

$$b = 3-4(1-\cos^2\alpha)=4\cos^2\alpha-1.$$

Now relate $$b+d$$ to $$b$$ and the unit length choice. Since $$b-d=1,$$ we have $$d=b-1,$$ so

$$b+d = b+(b-1)=2b-1.$$

But $$b+d$$ also equals $$2\cos\alpha,$$ therefore

$$2b-1 = 2\cos\alpha.$$

Insert $$b = 4\cos^2\alpha-1$$ from above:

$$2\left(4\cos^2\alpha-1\right)-1 = 2\cos\alpha.$$

Simplify step by step:

$$8\cos^2\alpha-2-1 = 2\cos\alpha$$

$$8\cos^2\alpha - 3 = 2\cos\alpha$$

$$8\cos^2\alpha - 2\cos\alpha - 3 = 0.$$

Let $$x=\cos\alpha.$$ The quadratic becomes

$$8x^2 - 2x - 3 = 0.$$

Apply the quadratic formula:

$$x = \frac{2 \pm \sqrt{2^2 - 4(8)(-3)}}{16}= \frac{2 \pm \sqrt{4+96}}{16}= \frac{2 \pm 10}{16}.$$

This gives two roots, $$x=\dfrac{12}{16}=\dfrac34$$ or $$x=-\dfrac12.$$ Because the smallest angle $$\alpha$$ must be acute, $$\cos\alpha>0,$$ so retain $$\cos\alpha=\dfrac34.$$

Consequently,

$$\sin\alpha = \sqrt{1-\left(\dfrac34\right)^2}= \sqrt{1-\dfrac9{16}}=\sqrt{\dfrac7{16}}=\dfrac{\sqrt7}{4}.$$

Now compute the side lengths relative to the unit smallest side.

Middle side:

$$b = 4\cos^2\alpha - 1 = 4\left(\dfrac34\right)^2 - 1 = 4\cdot\dfrac9{16}-1 = \dfrac9{4}-1 = \dfrac54.$$

Largest side:

$$b+d = 2b-1 = 2\left(\dfrac54\right)-1 = \dfrac{10}{4}-\dfrac4{4}= \dfrac6{4}=\dfrac32.$$

Hence, with $$b-d=1,\;b=\dfrac54,\;b+d=\dfrac32,$$ the proportional sides are

$$1 : \dfrac54 : \dfrac32.$$

Multiply every length by $$4$$ to clear denominators:

$$4 : 5 : 6.$$

Thus the three sides are in the ratio $$4:5:6,$$ which corresponds to Option D.

Hence, the correct answer is Option D.