Let the three angles be $a, b, c$ and WLOG assume $c \geq b \geq a$ Clearly $\sin b = \frac{\sin a + \sin c}{2}$ and $\cos ^2b = \cos a \cos c$ It is easy to get $$\sin ^2b = \frac{sin ^2a + \sin ^2c + 2\sin a \sin c}{4} = 1 - \cos a \cos c$$$$\boxed{4 - 4 \cos a \cos c} = \sin ^2a + \sin ^2c + 2 \sin a \sin c = \boxed{2 - \cos ^2a - \cos ^2c + 2 \sin a \sin c}$$This is equivalent to $$2 + (\cos a - \cos c)^2 = 2 \cos ({c - a}) \geq 2$$ Hence, $c = a$ and thus $a = b = c$

Let $x,y,z$ be those three angles,and let $\sin y$ be the middle term in the arithmetic sequence.Now,as sine and cosine,and geometric progressions are all monotonous in $[0,\frac{\pi}{2}]$,we deduce that $\cos y$ is the middle term in the geometric progression. Hence,$2\sin y=\sin x+\sin z=2\sin(\frac{x+z}{2})\cos(\frac{x-z}{2})$ and $\cos^2 y=\cos x\cdot\cos z=\frac{1}{2}(cos(x+z)+cos(x-z))\overset{\text{not}}{=}\frac{1}{2}(p+q)$. Now,$\sin^2 y=\frac{1-p}{2}\frac{1+q}{2}=\frac{1+q-p-pq}{4}$,thus $4=2(p+q)+1+q-p-pq\iff q(3-p)=3-p\iff q=1\iff x=z\iff x=y=z$. $\square$

Let $f(x) = ln \ cos \ arcsin \ x,\ x \in [0, 1).$ Assume that the angles are different and their sines $x_1, x_2, x_3$ form an arithmetic progression. Then the numbers $y_i = f(x_i)$ in the same order also form an arithmetic progression (because $f$ is monotone). Hence points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ of the Cartesian plane belong to the same line, and these points belong also to the graph of function $y = f(x).$ Therefore some line intersects the graph of $f(x)$ at three points. This is impossible because $f(x)$ is concave on the segment $[0, 1).$ Indeed, $f(x) = \frac{1}{2} ln (1-x^2), f'(x)= \frac{x}{x^2-1}, f''(x)= -\frac{x^2_1}{(x^2-1)^2} < 0.$