Question

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

Answer 1

`\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=-(1)/(2)\\ n=6 \end{cases}\implies a_6=(5)\left( -(1)/(2) \right)^(6-1) \\\\\\ a_6=(5)\left( -(1)/(2) \right)^(5)\implies a_6=(5)((-1)^5)/(2^5)\implies a_6=\cfrac{-5}{32}`