Given triangle ABC with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence. Find $\alpha, \beta, \gamma$. Given $\triangle ABC$ with $\alpha=2\beta$ and $b, c, a$ is forming an arithmetic sequence ($b$ is the $1$'st term, and $a$ is the last term), find $\alpha, \beta, \gamma$.
My attempts so far: 
Let $c=b+k$ and $a=b+2k$. Then, from $\alpha=2\beta$, I got $\gamma = 180^\circ - 3\beta$.
With the Law of Sines , I can write
$$\dfrac{b}{\sin \beta}=\dfrac{b+k}{\sin 3\beta}=\frac{b+2k}{\text{sin } 2\beta}$$
From $\dfrac{b}{\sin\beta}=\dfrac{b+k}{\sin 3\beta}$, it gives $\cos 2\beta = \dfrac{k}{2b}$.
From $\dfrac{b}{\sin \beta}=\dfrac{b+2k}{\sin 2\beta}$, it gives $\cos \beta = \dfrac{b+2k}{2b}$.
After this, I don't know what should I do. Is there any other theorem that can be used to solve this problem?

You can continue from where you left off. Using the fact that $\cos 2 \beta = 2 \cos^2 \beta - 1$:
$$2 \left( \frac{b+2k}{2b} \right)^2 - 1 = \frac{k}{2b}$$
$$2 (b + 2k)^2 - 4b^2 = 2bk$$
$$2b^2 + 8bk + 8k^2 - 4b^2 = 2bk$$
$$-2b^2 + 6bk + 8k^2 = 0$$
$$b^2 - 3bk - 4k^2 = 0$$
$$(b+k)(b-4k) = 0$$
but since $c = b + k \ne 0$, $b = 4k$ which gives side lengths $(4k, 5k, 6k)$ and has the same angles as $(4,5,6)$. Now the cosine rule $\cos C = \frac{a^2+b^2-c^2}{2ab}$ and so on can be used to find angles $\alpha,\beta,\gamma$.
