Question

What is the seventh term of the geometric sequence where a1=128 and a3=8

Answer 1

`\bf \begin{array}{lll} term&value\\ \cline{1-2} a_1&128\\ a_2&128r\\ a_3&128rr\\ &128r^2 \end{array}~\hspace{5em}\stackrel{a_3}{128r^2}=\stackrel{a_3}{8}\implies r^2=\cfrac{8}{128}\implies r^2=\cfrac{1}{16} \\\\\\ r=\sqrt{\cfrac{1}{16}}\implies r=\cfrac{√(1)}{√(16)}\implies r=\cfrac{1}{4}\qquad \leftarrow \textit{common ratio} \\\\[-0.35em] ~\dotfill`

`\bf n^(th)\textit{ term of a geometric sequence} \\\\ a_n=a_1\cdot r^(n-1)\qquad \begin{cases} a_n=n^(th)\ term\\ n=\textit{term position}\\ a_1=\textit{first term}\\ r=\textit{common ratio}\\ \cline{1-1} n=7\\ a_1=128\\ r=(1)/(4) \end{cases}\implies a_7=128\left( (1)/(4) \right)^(7-1) \\\\\\ a_7=128\left( (1)/(4) \right)^6\implies a_7=128\cdot \cfrac{1}{4096}\implies a_7=\cfrac{128}{4096}\implies a_7=\cfrac{1}{32}`