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?

Note that

r = 3/a

And the sum of the series can be represented as

S = 4 / [ 1 - 3/a ] = 4a / [ a - 3]

Note that if a = 4, then the sum is 4 (4) / [ 4 - 3] = 16/ 1 = 16 which is a perfect square