Question

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

Answer 1

`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`

Answer 2

`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`

Answer 3

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`