Question 419137
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:


<a href = "http://people.richland.edu/james/lecture/m116/sequences/geometric.html" target = "_blank">http://people.richland.edu/james/lecture/m116/sequences/geometric.html</a>


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