Find the sum, if it exists, of the infinite geometric series, 3 minus 3 over 5 plus 3 over 25 minus 3 over 125 plus etc

Question

Find the sum, if it exists, of the infinite geometric series,  

A	1
B
C
D	Sum does not exist

Answer 1

common ratio = r = -1/5 sum converges if -1<r<1. Eliminate answer D

a = 3

Sum = S = a/(1-r) = 3/(1+1/5) = 3/(6/5) = 3(5/6 = 5/2 = 2 1/2 = 2.5

B is the correct answer

the sum is 3 divided by 1+1/5 = 5/2

first 2 terms sum to 3-3/5 = 3-.6 = 2.4

first 3 terms sum to 2.4 + 3/25= 2.4 + 0.12 = 2.52

the remaining terms only change the sum by a tinier fraction. 2.4 and 2.52 are closest to 2.5. Sums of each new term will have the sum go slightly above 2.5 and then slightly below 2.5, oscillating. Answer is less than 3 which eliminates C.

Answer 2

You have a geometric series where a=3 and r = -1/5, so the usual sum of a geometric series expression applies: a/(1-r)