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The sum of a certain infinite geometric series is 2. The sum of the squares of all the terms is 3. Find the common ratio.

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The sum of a certain infinite geometric series is 2. The sum of the squares of all the terms is 3. Find the common ratio.

Jan 31, 2019

#4
+7825
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That means {$$\dfrac{a}{1-r} = 2\\ \dfrac{a^2}{1-r^2} = 3$$.

$$\dfrac{2a}{1+r} = 3\\ \dfrac{2(2-2r)}{1+r} = 3\\ 4-4r = 3+3r\\ 7r = 1\\ r = \dfrac{1}{7}\\ \dfrac{2a}{\frac{8}{7}} = 3\\ 7a = 12\\ a =\dfrac{12}{7}\\$$

So $$(r,a) = \left(\dfrac{1}{7},\dfrac{12}{7}\right)$$

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Feb 1, 2019

#1
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The formula for the sum of an infinite geometric sequence is a/1-r

1. Take that formula and make an equation from the problem.

"The sum of a certain infinite geometric series is 2"

So 2=a/1-r               (where r represents common ratio and a represents the first term

Then make another equation for the second part of the problem

"The sum of the squares of all the terms is 3"

So 3=(a^2)/1-(r^2)

this is true because when all the terms are squared, the common ratio is also squared.

2. Solve for the equations

Since we are trying to find the common ratio, solve for r.

We get 1/7 as the answer

Im not sure if this correct, will someone please check this, I hope this is not a bogus solution!

Jan 31, 2019
#3
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You have to set up 2 simultaneous equations as follows:

3=F^2 / (1 - r^2), 2 = F / (1 - r), solve for r, F

This gives r =1/7 and F =12/7

Guest Feb 1, 2019
#2
+1

YES!. It is correct. The common ratio is 1/7 and the first term 1 5/7 which all sum up to 2. And their squares sum up to 3. Here are the first 10 terms:

(1.714285714, 0.2448979592, 0.03498542274, 0.004997917534, 0.0007139882192, 0.000101998317, 0.00001457118815, 0.000002081598307, 0.000000297371186 7, 0.000000042481598 1, 0.000000006068799 728)

Jan 31, 2019
#4
+7825
+3
That means {$$\dfrac{a}{1-r} = 2\\ \dfrac{a^2}{1-r^2} = 3$$.
$$\dfrac{2a}{1+r} = 3\\ \dfrac{2(2-2r)}{1+r} = 3\\ 4-4r = 3+3r\\ 7r = 1\\ r = \dfrac{1}{7}\\ \dfrac{2a}{\frac{8}{7}} = 3\\ 7a = 12\\ a =\dfrac{12}{7}\\$$
So $$(r,a) = \left(\dfrac{1}{7},\dfrac{12}{7}\right)$$