Question

Determinar o nono termo da P.G. (1/2,-1,...)?

Answer 1

$a_{1}=\dfrac{1}{2},\ a_{2}=-1\\ \\ \\ \dfrac{a_{2}}{q}=a_{1}\Rightarrow \dfrac{-1}{q}=\dfrac{1}{2}\Rightarrow q\cdot 1=2\cdot -1\Rightarrow q=-2\\ \\ a_{n}=a_{1}\cdot q^{n-1}\\ \\ a_{9}=\dfrac{1}{2}\cdot \left(-2\right)^{9-1}=\dfrac{1}{2}\cdot \left(-2\right)^{8}=\dfrac{1}{2}\cdot 256=\dfrac{256}{2}=128$