Question

Find the sum to infinity of the infinite G.P.
1-½+¼........​

Answer 1

2/3

Explanation:

1-1/2+1/4.....

`r=((-1)/(2) )/(1) =-(1)/(2) \\if~|r|<1,then~s_(\infty)=(a)/(1-r) ,a=first~term.\\s_(\infty)=(1)/(1-((-1)/(2)) ) =(1)/(1+(1)/(2) ) =(1)/((3)/(2) ) =(2)/(3)`

Answer 2

`(2)/(3)`

Explanation:

The sum to infinity of a GP is

`S_(infinity)` =
`(a)/(1-r)` ; | r | < 1

where a is the first term and r the common ratio

Here a = 1 and r = -
`(1)/(2)` ÷ 1 = -
`(1)/(2)` , thus

`S_(infinity)` =
`(1)/(1-(-(1)/(2)) )` =
`(1)/(1+(1)/(2) )` =
`(1)/((3)/(2) )` =
`(2)/(3)`