#### Answer

The first five terms are:
$a_1 = \dfrac{1}{4}
\\a_2 = \dfrac{1}{16}
\\a_3 = \dfrac{1}{64}
\\a_4=\dfrac{1}{256}
\\a_5=\dfrac{1}{1024}$
The sequence is geometric with a common ratio of $\dfrac{1}{4}$.
The $n^{th}$ term is :$a_n = \dfrac{1}{4} \cdot\left(\dfrac{1}{4}\right)^{n-1}$

#### Work Step by Step

To find the first five terms, substitute 1, 2, 3, 4, and 5 to the given formula to obtain:
$a_1 = \dfrac{1}{4^1} = \dfrac{1}{4}
\\a_2 = \dfrac{1}{4^2} = \dfrac{1}{16}
\\a_3 = \dfrac{1}{4^3} = \dfrac{1}{64}
\\a_4=\dfrac{1}{4^4} = \dfrac{1}{256}
\\a_5=\dfrac{1}{4^5} = \dfrac{1}{1024}$
RECALL:
A sequence is geometric if there is a common ratio among consecutive terms.
Notice that $\dfrac{1}{4}$ times a term is equal to the value of the next term.
This means that the sequence has a common ratio of $\frac{1}{4}$.
Thus, the sequence is geometric with a common ratio of $\dfrac{1}{4}$.
The $n^{th}$ term of a geometric sequence is given by the formula $a_n = ar^{n-1}$
where $a$ = first term and $r$= common ratio.
This means that the $n^{th}$ term of the given geometric sequence, whose $a=\dfrac{1}{4}$ and $r=\dfrac{1}{4}$, is:
$a_n = \dfrac{1}{4} \cdot\left(\dfrac{1}{4}\right)^{n-1}$