Respuesta :

nap14hockey
Same as before, to make the formula do

t(1) * (r)^n-1
So 5 * (-1/2)^n-1
Therefore 
5*(-1/2)^5
0.078125
[tex]\bf n^{th}\textit{ term of a geometric sequence}\\\\ t_n=t_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ t_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ n=6\\ t_1=5\\ r=-\frac{1}{2}\\ \end{cases}\implies t_6=5\left(-\frac{1}{2} \right)^{6-1}[/tex]

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