SOLUTION: I hope someone can help me also with this question, "The sum of an infinite geometric series is twice its first term. Find the common ratio of the series." This question has been e

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Question 419137: I hope someone can help me also with this question, "The sum of an infinite geometric series is twice its first term. Find the common ratio of the series." This question has been eating at my brain for quite some time now. I hope you can help. Thank you.
Found 2 solutions by ewatrrr, Theo:
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!

Hi
"The sum of an infinite geometric series is twice its first term.
Find the common ratio of the series
= when |r| < 1
a/(1-r) = 2a
1/(1-r) = 2
1 = 2 - 2r
2r = 1
r = 1/2
In general: the sum of a geometric series is:

As the 'infinite' sum being just twice it's first term, was safe, in my mind,
to assume that r was a fraction and r^n would become so...insignificant that (1-r^n)= 1
For ex: (1/2)^20 = .0000001 demonstrates that processs

Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the formula for the sum of the first n terms of a geometric series is equal to:

S[n] = (a[1] * (1-r^n)) / (1-r) where a[1] is the first term in the series.

an example:

n = 3 and r = .7 and a[1] = 500

S[3] = (500 * (1-.7^3)) / (1-.7)

this becomes equal to 1095.

We can verify this by adding the elements together.

the nth element in a geometric series is given by the equation a[n] = a[1]*r^(n-1).

a[1] = 500 * .7^0 = 500
a[2] = 500 * .7^1 = 350
a[3] = 500 * .7^2 = 245

sum is 500 + 350 + 245 = 1095

as n approaches infinity, the formula of:

S[n] = (a[1] * (1-r^n)) / (1-r) where a[1] is the first term in the series.

becomes:

S[infinity] = a[1] / (1-r) where a[1] is the first term in the series.

r^n approaches 0 as n approaches infinity because r has to be smaller than 1.

this makes the expression (1-r^n) approach 1 as n approaches infinity.

an example:

let r = .9

.9^10 = .35
.9^100 = .000027
.9^1000 = 1.75 * 10^-46
.9^10000 = 0 on my calculator. it's not really 0 but it's so small that the calculator is not able to show it so it rounds the answer to 0.

at any rate, the formula for the sum of a geometric series as n approaches infinity is equal to:

S[infinity] = a[1] / (1-r) where a[1] is the first term in the series.

your problem states:
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The sum of an infinite geometric series is twice its first term. Find the common ratio of the series.
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if the sum of the infinite geometric series is twice its first term, then we get:

S[infinity] = 2*a[1]

we can replace S[infinity] in the equation of:

S[infinity] = a[1] / (1-r) where a[1] is the first term in the series.

by 2*a[1] to get:

2 * a[1] = a[1] / (1-r)

if we multiply both sides of this equation by (1-r) and we divide both sides of this equation by 2 * a[1] then we get:

(1-r) = a[1] / (2 * a[1])

we simplify this to get:

(1-r) = 1/2

we add r to both sides of this equation and we subtract 1/2 from both sides of this equation to get:

r = 1/2

let's see if this can be confirmed.

we assume r = 1/2 as we just calculated.

we let a[1] be any number chosen at random.

let's try 1469

S[infinity] = a[1] / (1-r) = 1469 / (1-(1/2) = 1469 / (1/2) = 2938

2938 is twice as large as 1469 so the equation is good.

it will work in all equations, since the general form would be:

S[infinity] = a[1] / (1 - (1/2) which always winds up being:

S[infinity] = a[1] / (1/2).

reference for formulas used is in the following link:

http://people.richland.edu/james/lecture/m116/sequences/geometric.html

Remember that r has to be smaller than 1 if we want to get the sum of the geometric series as n approaches infinity.