Find the term of a geometric sequence Precalculus

u_{n}=4\cdot \left( -\dfrac{2}{3} \right)^{n-1}

u_{n}=4\cdot \left( -\dfrac{32}{3} \right)^{n-1}

u_{n}=-6\cdot \left( -\dfrac{2}{3} \right)^{n-1}

u_{n}=4-\dfrac{2}{3}n

u_{n}=2\cdot \left(3 \right)^{n}

u_{n}=2\cdot \left(3 \right)^{n-1}

u_{n}=4n-2

u_{n}=\dfrac{2}{3}\cdot \left(3 \right)^{n-1}

u_{n}=8\cdot \left( -\dfrac{3}{4} \right)^{n}

u_{n}=6-\dfrac{3}{4}n

u_{n}=-6\cdot \left( \dfrac{3}{4} \right)^{n-1}

u_{n}=6\cdot \left( -\dfrac{3}{4} \right)^{n-1}

u_{n}= \left( \dfrac{1}{2} \right)^{n-1}

u_{n}= \dfrac{1}{2}\left( 2\right)^{n}

u_{n}= \left( \dfrac{1}{2} \right)^{n}

u_{n}=4\cdot \left( \dfrac{1}{2} \right)^{2n}

u_{n}= 6\cdot\left( -2 \right)^{n-1}

u_{n}= 24-18n

u_{n}= -6\cdot\left( 2 \right)^{n-1}

u_{n}= 3\cdot\left( 2 \right)^{n}

u_{n}= -4\cdot\left( -\dfrac{5}{4} \right)^{n}

u_{n}= 4\cdot\left( -\dfrac{5}{4} \right)^{n-1}

u_{n}=\dfrac{5}{4}n-4

u_{n}= -4\cdot\left( \dfrac{5}{4} \right)^{n-1}

u_{n}= 81\cdot\left( \dfrac{1}{3} \right)^{n}

u_{n}= 27\cdot\left( \dfrac{2}{3} \right)^{n-1}

u_{n}= 81\cdot\left( \dfrac{3}{2} \right)^{n-1}

u_{n}= 27-\dfrac{2}{3}n