SOLUTION: The nth term of a geometric term is 3^-n, the unlimited sum of the sequence is?
Can you please help me to solve this, I have no idea how to. Thank you!
Algebra ->
Sequences-and-series
-> SOLUTION: The nth term of a geometric term is 3^-n, the unlimited sum of the sequence is?
Can you please help me to solve this, I have no idea how to. Thank you!
Log On
Question 764513: The nth term of a geometric term is 3^-n, the unlimited sum of the sequence is?
Can you please help me to solve this, I have no idea how to. Thank you! Answer by ramkikk66(644) (Show Source):
nth term is given is 3^-n or 1/3^n. So the first term is 1/3, the second is 1/3^2
and so on.
The series is: 1/3,1/9,1/27... to infinity.
It is an infinite geometric progression with 1st term a as 1/3 and the common
ratio r as 1/3.
The formula for the sum to infinity of such a progression is given by
S = a / (1 - r) (I'm not including the proof for this here)
So here it is
The answer is
:)
(