How do you find the sum of the infinite geometric series given #18-12+8-...#?

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Answer 1 · 2018-06-24T07:51:29

#54/5#

for a GP

#a,ar,ar^2,...ar^(n-1),.....#

the sum to in infinity is

#S_oo=a/(1-r)#

provided
#|r|<1#

we have

#18-12+8-...#

#r=-12/18=-2/3#

#|-2/3|<1, :. EES_oo#

#S_oo=18/(1- -2/3)#

#=18/((1+2/3)#

#=18/(5/3)#

#=(18xx3)/5#

#=54/5#