1. The problem is to find the 12th term of the geometric progression (GP) given the first three terms: $1+i$, $2i$, $-2+2i$. 2. Recall that the $n$-th term of a GP is given by $$a_n = a_1 \cdot r^{n-1}$$ where $a_1$ is the first term and $r$ is the common ratio. 3. First, find the common ratio $r$ by dividing the second term by the first term: $$r = \frac{2i}{1+i}$$ 4. Simplify $r$: $$r = \frac{2i}{1+i} \cdot \frac{1 - i}{1 - i} = \frac{2i(1 - i)}{(1+i)(1 - i)} = \frac{2i - 2i^2}{1 - i^2}$$ Since $i^2 = -1$, $$2i - 2(-1) = 2i + 2 = 2 + 2i$$ and denominator $$1 - (-1) = 2$$ So, $$r = \frac{2 + 2i}{2} = 1 + i$$ 5. Verify $r$ by confirming the third term: $$a_3 = a_1 \cdot r^{2} = (1 + i)(1 + i)^{2} = (1 + i)^3$$ Calculate $(1 + i)^3$: $$(1+i)^2 = 1 + 2i + i^2 = 1 + 2i -1 = 2i$$ Then $$(1+i)^3 = (1+i)^2 (1+i) = 2i (1+i) = 2i + 2i^2 = 2i - 2 = -2 + 2i$$ This matches the given $a_3$. 6. Now, find the 12th term: $$a_{12} = a_1 \cdot r^{11} = (1+i)(1+i)^{11} = (1+i)^{12}$$ 7. Use De Moivre's Theorem to simplify $(1+i)^{12}$: Convert $1+i$ to polar form: $$|1+i| = \sqrt{1^2 + 1^2} = \sqrt{2}$$ $$\theta = \arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}$$ So, $$(1+i)^{12} = (\sqrt{2})^{12} (\cos(12 \times \pi/4) + i \sin(12 \times \pi/4)) = (\sqrt{2})^{12} (\cos(3\pi) + i \sin(3\pi))$$ Since $$(\sqrt{2})^{12} = (2^{1/2})^{12} = 2^{6} = 64$$ and $$\cos(3\pi) = -1$$ $$\sin(3\pi) = 0$$ Therefore, $$a_{12} = 64 (-1 + 0i) = -64$$ Final answer: The 12th term of the geometric progression is $\boxed{-64}$.