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## Question:

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

## Answer:

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

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