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Let
a + ar + ar^2 + ar^3 + \dotsb
be an infinite geometric series. The sum of the series is $4.$ The sum of the cubes of all the terms is $10.$ Find the common ratio.

 Apr 11, 2024
 #1
avatar+33616 
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\(a+ar+ar^2+...=\frac{a}{1-r}\) as long as r<1

 

\(a^3+a^3(r^3)+a^3(r^3)^2+...=\frac{a^3}{1-r^3}\)  as long as r^3<1

 

So:  

\(\frac{a}{1-r}=4\)

and

\(\frac{a^3}{1-r^3}=10\)

 

Can you take it from here?

 Apr 12, 2024

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