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Consider the geometric series \(4+\frac{12}{a}+\frac{36}{a^2}+\cdots\). If the sum is a perfect square, what is the smallest possible value of a where a is a positive integer?

 Sep 25, 2016

Best Answer 

 #1
avatar+36915 
+10

Since a geometric series can be expressed as   a1 + a1 r + a1r^2 + a1 r^3

we can see that  r = 3/a  from the terms given

the SUM of a geometric series is  S = a1 / (1-r)

S = a1 /(1-r)

  = 4/(1-r)   needs to be a perfct square....

by brute force solving this equation with S= perfect square , '9' will not work , but '16' will

 

16 = 4/ (1- 3/a)

16- 48/a = 4

48/a = 12

a = 4                   Correct???

 Sep 29, 2016
 #1
avatar+36915 
+10
Best Answer

Since a geometric series can be expressed as   a1 + a1 r + a1r^2 + a1 r^3

we can see that  r = 3/a  from the terms given

the SUM of a geometric series is  S = a1 / (1-r)

S = a1 /(1-r)

  = 4/(1-r)   needs to be a perfct square....

by brute force solving this equation with S= perfect square , '9' will not work , but '16' will

 

16 = 4/ (1- 3/a)

16- 48/a = 4

48/a = 12

a = 4                   Correct???

ElectricPavlov Sep 29, 2016
 #2
avatar+118587 
+10

Consider the geometric series \(4+\frac{12}{a}+\frac{36}{a^2}+\cdots\)

 If the sum is a perfect square, what is the smallest possible value of a where a is a positive integer?

 

This is a GP where  T1=4     and      r= 3/a

 

 \(S_{\infty}=\frac{T_1}{1-r}=\frac{4}{1-\frac{3}{a}}=4\div \frac{a-3}{a}=\frac{4a}{a-3}\)

 

Now this sum is a perfect square so the square root of it is a rational number so

let's look at this.

 

\(\sqrt{\frac{4a}{a-3}}\\ =\frac{\sqrt{4a}}{\sqrt{a-3}}\\ =\frac{2\sqrt{a}}{\sqrt{a-3}}\\ \)

 

So a has to be a perfect square and a-3 has to be a perfect square.

The smallest value of a to meet this criterion is a=4

 

I have just seen your answer ElectricPavlov.  YES your answer is correct :))

 Sep 29, 2016

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