What is the sum of the infinite geometric series?

-3-3/2-3/4-3/8-3/16-

a. - 93/16

b. 3/32

c. - 4

d. - 6

That is a geometric sequence of the form a (n) = - 3 (1/2) ^ (n-1)

The sum of any geometric sequence is:

s (n) = a (1-r^n) / (1-r)

and if r^2<1 it converges to a sum of:

s=a / (1-r) ... in this case:

s=-3 / (1-1/2)

s=-3 / (1/2)

s=-6