geometric progression

Nice. 

$\frac{a}{1-r}=3 \rightarrow a=3-3r$ by geometric sum formula, where r is the common ratio. Our sequence would be $a,ar,ar^2,ar^3,...$. When squared, it would be $a^2, a^2r^2, a^2r^4, a^2r^6$. This sequence has first term a^2 and common ratio r^2, so by the geometric sum formula this is $\frac{a^2}{1-r^2}=3 \rightarrow \frac{(3-3r)^2}{1-r^2}=3\rightarrow 9r^2-12r+9=3-3r^2 \rightarrow r=\frac{1\pm i}{2}$. So also $a=3-\frac{3\pm 3i}{2}=\frac{3\pm 3i}{2}$.