How do you find the sum of the infinite geometric series 27+9+3+1+...?

3 Answers
Harish Chandra Rajpoot
Jul 11, 2018

#40.5#

Explanation:

The given infinite geometric series

#27+9+3+1+\ldots#

has first term #a=27# & the common ratio #r=9/27=3/9=\ldots=1/3# h

Hence the sum of given infinite G.P.

#=a/{1-r}#

#=\frac{27}{1-1/3}#

#=81/2#

#=40.5#

Jim G.
Jul 11, 2018

#S_oo=81/2#

Explanation:

#"the sum to infinity of a geometric series is"#

#•color(white)(x)S_oo =a/(1-r)to -1 < r <1#

#"where a is the first term and r the common ratio"#

#"here "a=27" and "r=9/27=3/9=1/3#

#S_oo=27/(1-1/3)=27/(2/3)=81/2#

Narad T.
Jul 11, 2018

The answer is #=40.5#

Explanation:

If the common ratio of a geometric series is

#|r|<1#

Then,

The sum of the infinite geometric series

#a_1+a_1r+a_1r^2+........a-1r^n+.......#

is given by .

#S_(oo)=a_1/(1-r)#

Here the series is

#27+ 27*1/3+27*(1/3)^2+27*(1/3^3)+.................#

#a_1=27#

and

#r=1/3#

#S_(oo)=27/(1-1/3)=27/(2/3)=40.5#

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