To: Max.....

Sigh...... Already tried that few days ago...... Me not good at G.S. :(

I think Melody may know how to do this question. 

Melody is much better at these questions than me, but here is my 'guess'

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???

I answered on the original question.

and yes EP is correct :))

4 + 12/a + 36/a^2 + 108/a^3 + 324/a^4.................If a=4, then by convergence test we get:

[4+∑(4*3^n)/4^n)), n=1 to ∞] =16, so a=4 is correct. The infinite series: 4/(1 - r)=16, r=3/4.

Source: http://www.wolframalpha.com/input/?i=%5B4%2B%E2%88%91(4*3%5En)%2F4%5En)),+n%3D1+to+%E2%88%9E%5D