Answer
$a_n = \sqrt2 \cdot (\sqrt2)^{n-1}$
(This is equivalent to $a_n=(\sqrt{2})^{n}$.)
$a_5=4\sqrt2$
Work Step by Step
RECALL:
The $n^{th}$ term $a_n$ of a geometric sequence is given by the formula:
$a_n=a_1 \cdot r^{n-1}$
where
$a_1$ = first term
$r$ = common ratio
The given geometric sequence has:
$a_1=\sqrt2$
$r=\sqrt2$
Thus, the $n^{th}$ term of the sequence is given by the formula:
$a_n = \sqrt2 \cdot (\sqrt2)^{n-1}$
(This is equivalent to $a_n=(\sqrt{2})^{n}$.)
The 5th term can be found by substituting $5$ for $n$:
$a_5=\sqrt2 \cdot \sqrt2)^{5-1}
\\a_5=\sqrt2 \cdot (\sqrt2)^4
\\a_5 = \sqrt2 \cdot 4
\\a_5=4\sqrt2$