1. **Problem:** Determine whether the 12th term of the sequence $1+i, 2i, -2+2i, \dots$ is $10+12i$. 2. **Identify the type of sequence:** Observe the terms: - Term 1: $1+i$ - Term 2: $2i$ - Term 3: $-2+2i$ Let's check if it is an arithmetic sequence by finding the common difference $d$. 3. **Calculate common differences:** \[ d_1 = \text{Term }2 - \text{Term }1 = 2i - (1+i) = -1 + i \] \[ d_2 = \text{Term }3 - \text{Term }2 = (-2+2i) - 2i = -2 \] Since $d_1 \neq d_2$, this is not an arithmetic sequence. 4. **Test for geometric sequence:** Find ratio $r$: \[ r = \frac{\text{Term }2}{\text{Term }1} = \frac{2i}{1+i} \times \frac{1-i}{1-i} = \frac{2i(1 - i)}{(1+i)(1 - i)} = \frac{2i - 2i^2}{1 - i^2} = \frac{2i - 2(-1)}{1 - (-1)} = \frac{2i + 2}{2} = 1 + i \] Check if Term 3 matches Term 2 times $r$: \[ (2i)(1+i) = 2i + 2i^2 = 2i - 2 = -2 + 2i \] This matches Term 3. 5. **General term of geometric sequence:** \[ a_n = a_1 \times r^{n-1} = (1+i)(1+i)^{n-1} = (1+i)^n \] 6. **Calculate the 12th term:** \[ a_{12} = (1+i)^{12} \] 7. **Simplify $(1+i)^{12}$:** Note that $1+i = \sqrt{2}e^{i\pi/4}$. So, \[ (1+i)^{12} = (\sqrt{2})^{12} e^{i \frac{12 \pi}{4}} = (\sqrt{2})^{12} e^{i3\pi} = (\sqrt{2})^{12} (\cos 3\pi + i \sin 3 \pi) \] \[ (\sqrt{2})^{12} = (2^{1/2})^{12} = 2^{6} = 64 \] \[ \cos 3\pi = -1, \quad \sin 3\pi = 0 \] Therefore, \[ (1+i)^{12} = 64 (-1) = -64 + 0i = -64 \] 8. **Compare with given term:** The 12th term is $-64$, which is not $10+12i$. **Final answer:** No, the 12th term of the sequence is $-64$, not $10+12i$.