Given:
- The lengths of the sides of a triangle are in arithmetic progression (AP).
- The greatest angle is double the smallest angle.

To find:
The ratio of the lengths of the sides of the triangle.

Solution:

Let the lengths of the sides of the triangle be a-d, a, and a+d, where a is the middle term of the AP and d is the common difference.

Step 1: Find the angles:

Let A, B, and C be the angles of the triangle corresponding to the sides a-d, a, and a+d respectively.

According to the given condition, the greatest angle C is double the smallest angle A.

So, C = 2A

Also, the sum of all angles in a triangle is 180 degrees.

A + B + C = 180

Substituting the value of C, we get:

A + B + 2A = 180
3A + B = 180

Step 2: Use the sine rule:

The sine rule states that for any triangle with sides a, b, and c opposite to angles A, B, and C respectively:

a/sinA = b/sinB = c/sinC

In this case, we have:

(a-d)/sinA = a/sinB = (a+d)/sinC

Using the sine rule, we can rewrite the equation:

(a-d)/sinA = a/sinB = (a+d)/sinC = k (let)

Step 3: Solve the equations:

From the equation A + B + 2A = 180, we can solve for B:

3A + B = 180
B = 180 - 3A

Substituting the values of A and B in the equation (a-d)/sinA = a/sinB = (a+d)/sinC = k, we get:

(a-d)/sinA = a/sin(180-3A) = (a+d)/sin2A

Simplifying the equation, we get:

(a-d)/sinA = a/sin(180-3A) = (a+d)/sin2A
(a-d)/sinA = a/sin3A = (a+d)/sin2A

Cross-multiplying, we get:

(a-d)sin3A = asinA
(a+d)sinA = asin2A

Expanding and simplifying, we get:

3asin3A - dsin3A = asinA
asin2A + dsinA = asinA

Dividing these two equations, we get:

(3asin3A - dsin3A)/(asin2A + dsinA) = 1

Simplifying further, we get:

3sin3A - dsin3A = asin2A + dsinA

Using the trigonometric identity sin3A = 3sinA - 4sin^3A and sin2A = 2sinAcosA, we can rewrite the equation as:

3(3sinA - 4sin^3A) - d(3sinA - 4sin^3A) = a(2sinAcosA) + dsinA

Expanding and simplifying, we get:

9sin