When I got this question, my first thought was a triangle with three angles of 30, 60 and 90 respectively, but in the end it was an equilateral triangle.

Solution: Because A, B and C become arithmetic progression and A, B and C become geometric progression, it is assumed that:

B=A+Q，C=A+2Q

b = c=aq^2 AQ

And a+b+c = 180, that is, 3a+3q = 180, so:

B angle: a+q = 60.

A+B= 120

According to sine theorem, a/b = Sina/sinb and b/c = sinb/sinc, namely:

1/q=sinA/sin60， 1/q = sin 60/sin( 120-A)

From the above formula, we can get:

sinAsin( 120 -A)=(sin60 )^2

Simplification: √ 3/2sinacosa+1/2sinasina = 3/4.

√3/2 sin2a- 1/2 cos2a = 1

sin(2A-30 )=sin90

Finally, a = 60 is obtained.

So triangle ABC is an equilateral triangle. And its Q=0, q= 1.