cuhmoon wrote:
Hi Bunuel,
Can we say that if |r| <1, then it's an infinite GP ? How does one define infinite GP?
When does one use the forumula
Sum = b1 (r^n-1)/r-1 ?
Please explain..
Bunuel wrote:
Stiv wrote:
\(\frac{first \ term}{1-constant}\) Is this formula reversed when we have an increase by 0<constant<1? Does it look like this \(\frac{first \ term}{1+constant}\)?
The sum of
infinite geometric progression with
common ratio \(|r|<1\), is \(sum=\frac{b}{1-r}\), where \(b\) is the first term.
To add to what Bunuel said, r is the common ratio and has nothing to do with whether a progression is infinite or finite. In either case, r can be less than 1 or more than 1
e.g.
1, 3, 9, 27, 81 ... (infinite sequence with r = 3)
27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3)
1, 3, 9 (finite sequence with 3 elements and r = 3)
27, 9, 3 (finite sequence with 3 elements and r = 1/3)
Now the point is that the sum of all terms of the first sequence is infinite. The terms will can getting larger and will keep adding up. So the sum will be infinite.
We can find the exact sum of the rest of the 3 sequences.
27, 9, 3, 1, 1/3, 1/9 ... (infinite sequence with r = 1/3)
Sum = a/(1 - r) = 27/(1 - 1/3) = 81/2
1, 3, 9 (finite sequence with 3 elements and r = 3)
Sum = a(r^n - 1)/(r - 1) = 1*(3^3 - 1)/(3 - 1) = 13
27, 9, 3 (finite sequence with 3 elements and r = 1/3)
Sum = a(1 - r^n)/(1 - r) = 27*(1 - 1/3^3)/(1 - 1/3) = 26*3/2 = 39
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