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(A) The sum is a geometric progression: $S = \sum_{n=1}^{2025} (-\omega)^n = (-\omega) + (-\omega)^2 + \dots + (-\omega)^{2025}$.This is a geometric s

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$-2$

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(A) The sum is a geometric progression: $S = \sum_{n=1}^{2025} (-\omega)^n = (-\omega) + (-\omega)^2 + \dots + (-\omega)^{2025}$.This is a geometric series with first term $a = -\omega$,common ratio $r = -\omega$,and $n = 2025$ terms.$S = \frac{a(1-r^n)}{1-r} = \frac{-\omega(1-(-\omega)^{2025})}{1-(-\omega)} = \frac{-\omega(1 - (-\omega^{2025}))}{1+\omega}$.Since $\omega^3 = 1$ and $2025$ is a multiple of $3$,$\omega^{2025} = 1$.$S = \frac{-\omega(1 - (-1))}{1+\omega} = \frac{-\omega(2)}{1+\omega}$.Using $1+\omega = -\omega^2$,we get $S = \frac{-2\omega}{-\omega^2} = \frac{2}{\omega} = 2\omega^2$.Since $\omega = e^{i2\pi/3}$,$\omega^2 = e^{i4\pi/3} = e^{-i2\pi/3}$.Thus,$\alpha = \arg(2\omega^2) = \arg(e^{-i2\pi/3}) = -\frac{2\pi}{3}$.Therefore,$\frac{3\alpha}{\pi} = \frac{3}{\pi} 	imes \left(-\frac{2\pi}{3}ight) = -2$.

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