+0  
 
0
56
1
avatar+397 

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?

 Apr 26, 2020
 #1
avatar+21017 
+1

Formula:  Sum  =  (first term) / [ 1 - (common ratio) ]

 

For this series, the common ratio is  3/a

 --->     Sum  =  4 / [ 1 - 3/a ]     --->     Sum  =  (4a) / (a - 3)

 

--->  4 is the smallest value for a to result in a perfect square   

 Apr 26, 2020

12 Online Users