The three consecutive terms of a geometric sequence  are the second, third and sixth term of an arithmetic sequence. 
     Find the common ratio of the GP.

1.  Our numbers (1-st, 2nd and 3-rd) are a, ar and ar^2, three consecutive terms of an geometric progression 
    with the first term "a" and the common ratio "r".


2.  Since a and ar are the consecutive terms of an arithmetic progression, their difference is "d", the common difference of the AP

    ar - r = d.       (1)


    Since ar and ar^2 are the second and the sixth terms of the AP, their difference is four times d": 

     = 4d.     (2)


3.  From (1) and (2) you have this equation

    4(ar-a) = ,   which is equivalent to

    4a*(r-1) = ar*(r-1).


    Assuming that  a=/=0 and r=/=1, we can cancel the factors "a" and (r-1) in both sides and to get

    4 = r.


    It is our answer: r = 4.


4.  Now let us consider these exclusive cases  a = 0  and  r= 1.


    a) if a = 0, then the three terms of the GP are 0, 0, and 0.

       This progression satisfies all the problem conditions, but it is not very interesting. It is a degenerated case. 
       But still satisfies all conditions.


    b) if r = 1, then the three terms of the GP are a, a, and a,  for any arbitrary "a". Considered as the arithmetic progression, 
       it has the common difference 0 (zero). 

       Its 6-th term is also "a", and such a set satisfies all the problem conditions, again. Although is degenerated, too.


5.  Therefore, the answer is: There is a unique non-degenerated solution with the common ratio 4.
                              There are infinitely many degenerated solutions with the common ratio 1.