If  k+1, 2k-1, 3k+1  are three consecutive terms of a geometric progression, then

    the ratio    is equal to the ratio  ,

by the definition of a geometric progression.


Hence,

     = .


It implies

   (3k+1)*(k+1) = (2k-1)^2

   3k^2 + k + 3k + 1 = 4k^2 - 4k + 1

   k^2   - 8k = 0

   k*(k-8) = 0

which has two roots  k= 0  and  k= 8.


If k= 0, then the first and the second terms of the GP are  k+1 = 1  and  2k-1 = -1, so the common ratio is   = -1.


If k= 8, then the first and the second terms of the GP are  k+1 = 9  and  2k-1 = 15, so the common ratio is   = .


ANSWER.  Under given conditions, the common ratio may have one of the two values  -1  and/or  .