Infinite GP

Let the common ratio of the GP be $r$ , then we have:

$\begin{aligned} a_1 \cdot a_2 \cdot a_3 \cdot a_4 & = 64 \\ a_1 \cdot a_1r \cdot a_1r^2 \cdot a_1r^3 & = 64 \\ a_1^4 r^6 & = 2^6 \end{aligned}$

For $a_1$ , $a_2$ , $a_3$ and $a_4$ to be integers, both $a_1$ and $r$ must also be integers. For $a_1^4 r^6 = 2^6$ , the solution must be $a_1 = 1$ and $r=2$ . Therefore, $a_1=1$ , $a_2=2$ , $a_3=4$ , $a_4=8$ and $a_5 = \boxed{16}$ .

let a1=x/r^3

a2=x/r

a3=xr

a4=x * r^3

a1 * a2 * a3 * a4= x^4

x^4=64

x=2*2^(1/2)

however a1,a3,a3,a4.... are positive integers

therefore r=2^(1/2)

therefore the gp is 1,2,4,8...

therefore a5 will be 16