Answer
$a_n=16\left(\dfrac{1}{4}\right)^{n-1}$
Work Step by Step
In the given geometric sequence, $16,4,1,...$, $a_1=16$ and $r=\dfrac{1}{4}$. Using $ a_n=a_1r^{n-1} $ or the formula for the $n^{th}$ term of a geometric sequence, then \begin{align*}\require{cancel} a_n&=16\left(\dfrac{1}{4}\right)^{n-1} .\end{align*} Hence, the general term, $a_n,$ of the geometric sequence $16,4,1,...$ is given by $a_n=16\left(\dfrac{1}{4}\right)^{n-1}$.
