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# Help again!

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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?

MIRB16  Sep 11, 2017
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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

CPhill  Sep 11, 2017

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