(1+i,\ 2i,\ -2+2i,\ \ldots)

[ r=\dfrac{2i}{1+i}=\dfrac{2i(1-i)}{2}=i(1-i)=i-i^2=1+i ]

Check: ((1+i)\cdot(1+i)=1+2i-1=2i) ✓

[ a_{12}=a r^{11}=(1+i)(1+i)^{11}=(1+i)^{12} ]

[ 1+i=\sqrt{2}e^{i\pi/4}\implies(1+i)^{12}=(\sqrt{2})^{12}e^{i3\pi}=64\cdot(-1)=-64 ]

Answer: (-64)