An infinite geometric series has first term -3/2 and sums to twice the common ratio. Find the sum of all possible values for the common ratio.

The formula for a convergent infinite geometric series is:  Sum  =  a / (1 - r)

where  a = first term  and  r = common ratio.

This problem has the first term = -3/2   --->   a = -3/2

Since it sums to twice the common ratio,  Sum = 2r.

Putting these together   --->   2r  =  (-3/2) / (1 - r)                        multiply both sides by 1 - r:

                                      --->   2r(1 - r)  =  -3/2                              multiply both sides by 2:

                                      --->   4r(1 - r)  =  -3                                 multiply out the left-hand side:

                                      --->   4r - 4r²  =  -3                                  rearrange the terms:

                                      --->   0  =  4r² - 4r - 3                              factor:

                                      --->   0  =  (2r - 3)(2r + 1)

                                      --->   Either  r = 3/2  or  r = -1/2

 

Since an infinite geometric series converges only when  -1 < r < r,  the possible answer 3/2 must be rejected.

             --->   answer:  r = -1/2