SOLUTION: the 3rd and 4th terms of geometric progression are 12 and 8. find the sum to infinity progression.
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Question 1082960: the 3rd and 4th terms of geometric progression are 12 and 8. find the sum to infinity progression.
Found 2 solutions by KMST, amfagge92:
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
A geometric progression (called a geometric sequence in the USA) is a sequence of numbers (called terms) where the ratio of consecutive terms is always the same. That ratio is called the common ratio, .
In this case .
If we call the first term of our sequence ,
the first terms of the progression are
, , , , ..., .
The sum of those terms is
.
When , , so we like to write it as
instead.
In that case, as increases, becomes smaller,
tending to zero.
So the sum to infinity is .
In this case, the third term is
, or , or .
So, , and the sum to infinity is
=== .
Answer by amfagge92(93) (Show Source): You can put this solution on YOUR website!
3rd term=12
4th term=8
Nth term=ar^n-1
12=ar^2..i
8=ar^3..ii
divide through to eliminate a
3/2=r^-1
Solving i have
r=2/3
⇒12=ar^2..from i
12=a*4/3
a=27
a=27
r=2/3
S∞=a/(1-r)
=81
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