Answer
$r = \sqrt2$
$a_5=4$
The $n^{th}$ term of the geometric sequence is: $a_n=1 \cdot \left(\sqrt2\right)^{n-1}$
Work Step by Step
RECALL:
(1) The common ratio of a geometric sequence is equal to the quotient of any two consecutive terms:
$r =\dfrac{a_n}{a_{n-1}}$
(2) The $n^{th}$ term of a geometric sequence is given by the formula:
$a_n = a\cdot r^{n-1}$
where
$a$ = first term
$r$ = common ratio
The sequence is said to be geometric.
Thus, we can proceed to solving for the common ratio:
$r=\dfrac{\sqrt2}{1}
\\r = \sqrt2$
The fifth term can be found by multiplying the common ratio to the fourth term.
The fourth term is $2\sqrt2$.
Thus, the fifth term is:
$a_5=2\sqrt2 \cdot \sqrt2
\\a_5=2(2)
\\a_5=4$
With a first term of $1$ and a common ratio of $r=\sqrt2$, the $n^{th}$ term of the geometric sequence is:
$a_n=a \cdot r^{n-1}
\\a_n=1 \cdot \left(\sqrt2\right)^{n-1}$