+0

# Geometric Series Problem

0
131
1
+84

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

benjamingu22  May 27, 2017
Sort:

#1
+76870
+1

The sum of this inifnite series  =

4 / ( 1 - 3/a)        where  4  is the first term  and  (3/a)  is the common ratio

Simplifying this, we have

4a /  [ a - 3]

Note.........that  since a is positive, the first value of a that produces a perfect square is when a  = 4

4(4) / [ 4 - 3 ]  =  16 / 1    =  16    which is a perfect square

CPhill  May 27, 2017

### 12 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details