Respuesta :
[tex]\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} a_5=\frac{1}{16}\\ r=\frac{1}{4}\\ n=5 \end{cases}\implies \cfrac{1}{16}=a_1\left(\cfrac{1}{4} \right)^{5-1}[/tex]
[tex]\bf \cfrac{1}{16}=a_1\left(\cfrac{1}{4} \right)^4\implies \cfrac{1}{16}=a_1\left( \cfrac{1^4}{4^4} \right)\implies \cfrac{1}{16}=a_1\cdot \cfrac{1}{256}\implies \cfrac{1}{16}=\cfrac{a_1}{256} \\\\\\ \cfrac{256}{16}=a_1\implies 16=a_1~\hfill \boxed{a_n=16\left( \frac{1}{4} \right)^{n-1}}[/tex]