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