How to show that this decision problem is in co-NP?
P is a subset both of NP and of co-NP. Do you think the problem is in P?

(O(n^3) solution is trivial if poster states the problem correctly. As it is, $i_1 = i_2 = i_3  ... = 1$ is a solution.)

You need n/2 different indices $i_j$ such that $a_{i_2} / a_{i_1} = a_{i_3} / a_{i_2}$, $a_{i_3} / a_{i_2} = a_{i_4} / a_{i_3}$, etc. 

There are $n^2 - n$ ways to choose $i_1$ and $i_2$. Once you chose these, let $d = a_{i_2} / a_{i_1}$. Then we must have $a_{i_3}$ = $a_{i_2} \cdot d$, $a_{i_4}$ = $a_{i_2} \cdot d^2$, $a_{i_5}$ = $a_{i_2} \cdot d^3$ etc. So all you have to do is check whether there are enough $a_i$ with these values and take their indices. 

You can do this faster, but all I cared about was whether it's polynomial time.