+249 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?
+37146 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???
+37146 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???
+118673 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 :))
