Let's consider a triangle $ABC$ with angles $A$, $B$, and $C$. By the given condition that the angles form an arithmetic progression, we can write the following relation:
$$2B = A + C.$$

By knowing that the sum of interior angles in any triangle is 180°, we have:
$$A + B + C = 180°.$$

Substituting $2B$ for $A + C$ from the arithmetic progression condition into this equation, we get:
$$A + 2B + C = 180° \\ 2B + 2B = 180° \\ 4B = 180° \\ B = \dfrac{180°}{4} \\ B = 45°.$$

This is incorrect as $B$ should be 60° based on the premise of the problem. It seems there has been a mistake in the translation or interpretation of the initial solution.

Let's attempt to correct this:

From $2B = A + C$ and knowing that $A, B,$ and $C$ are in an arithmetic progression, we have:
$$B - A = C - B.$$
By substitution, this becomes:
$$B - A = C - (180° - A - C) \\ B - A = C - 180° + A + C \\ 2B = 2A + 2C - 180° \\ 2B = 180° \\ B = 90°.$$

This must be revisited as we know no angle in a triangle can be 90° given the other two angles must also have nonzero values. Let's correct the arithmetic accordingly.

Combining the arithmetic progression relationship and the sum of angles, we have:
$$B-A = C-B \text{ and } A+B+C=180°.$$
Adding these two equations, we get:
$$A+2B+C=2B+180° \\ 2A+2C=180°.$$

Dividing this equation by 2, we find that:
$$A+C=90°.$$

Substituting it back to our arithmetic progression relation, we get:
$$2B=A+C=90°,$$
and therefore,
$$B = \dfrac{90°}{2} = 45°,$$

which is also incorrect as it conflicts with the given statement that one of the angles is 60°.

The correct approach should be to use the arithmetic progression relations directly. If the angles are in arithmetic progression, say $A = B - x$ and $C = B + x$ for some common difference $x$. So we have:
$$A + B + C = (B - x) + B + (B + x) = 3B = 180°,$$
resulting in the conclusion that:
$$B = \frac{180°}{3} = 60°.$$
This asserts that angle $B$ is 60°.

Now let's consider the geometric progression of sides. Let the sides of the triangle $ABC$ opposite to the angles $A$, $B$, and $C$ be $a$, $b$, and $c$, respectively, in geometric progression. This implies:
$$\frac{a}{b} = \frac{b}{c} \implies b^2 = ac.$$

Using the law of cosines for angle $B$, we have:
$$b^2 = a^2 + c^2 - 2ac\cos(B),$$
and since we have established $B = 60°$, and $\cos(60°) = \frac{1}{2}$, the equation simplifies to:
$$b^2 = a^2 + c^2 - ac.$$

From the condition of geometric progression $b^2 = ac$, and replacing it in the above equation, we get:
$$ac = a^2 + c^2 - ac,$$
and this simplifies to:
$$0 = a^2 - 2ac + c^2 = (a - c)^2.$$

Therefore, we have $a - c = 0$, meaning $a = c$. Since we have $a = c$ and angle $B$ is 60°, by the properties of triangles, we can conclude that:
$$A = C = 60°.$$

Thus, triangle $ABC$ is equilateral, and all of its angles are 60°. We can state our final result as:
$$\boxed{A = B = C = 60°}.$$