If theta = frac{pi}{6},then the 10^{th} term | vedclass

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(D) The given series is a geometric progression with the first term $a = 1$ and common ratio $r = (\cos 	heta + i \sin 	heta)$.The $n^{th}$ term of

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$-i$

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(D) The given series is a geometric progression with the first term $a = 1$ and common ratio $r = (\cos 	heta + i \sin 	heta)$.The $n^{th}$ term of a geometric progression is given by $T_n = a \cdot r^{n-1}$.For the $10^{th}$ term $(n = 10)$,we have $T_{10} = 1 \cdot (\cos 	heta + i \sin 	heta)^{10-1} = (\cos 	heta + i \sin 	heta)^9$.Using De Moivre's theorem,$(\cos 	heta + i \sin 	heta)^n = \cos(n	heta) + i \sin(n	heta) = e^{i n 	heta}$.Substituting $	heta = \frac{\pi}{6}$ and $n = 9$:$T_{10} = e^{i 9 (\frac{\pi}{6})} = e^{i \frac{3\pi}{2}}$.Using Euler's formula $e^{i \phi} = \cos \phi + i \sin \phi$:$T_{10} = \cos(\frac{3\pi}{2}) + i \sin(\frac{3\pi}{2}) = 0 + i(-1) = -i$.

If $\theta = \frac{\pi}{6}$,then the $10^{th}$ term of the series $1 + (\cos \theta + i \sin \theta) + (\cos \theta + i \sin \theta)^2 + (\cos \theta + i \sin \theta)^3 + \ldots$ is equal to:

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