Marilynlk
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# Find the 6th term of a geometric sequence with t1=5 and r = -1/2

$\bf n^{th}\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\ --------\\ a_1=5\\ r=-\frac{1}{2}\\ n=6 \end{cases}\implies a_6=(5)\left( -\frac{1}{2} \right)^{6-1} \\\\\\ a_6=(5)\left( -\frac{1}{2} \right)^{5}\implies a_6=(5)\frac{(-1)^5}{2^5}\implies a_6=\cfrac{-5}{32}$