Answer
$9\sqrt2.$
Work Step by Step
We know that $a_n=a_1+(n-1)d.$
where $d$=common difference and $a_1$= the first term
Here $a_1=\sqrt2$ and $d=a_2-a_1=2\sqrt2-\sqrt2=\sqrt2$
Hence, here we have
$a_n=\sqrt2+(n-1)\cdot\sqrt2.$
Therefore $a_{9}=\sqrt2+(9-1)\cdot\sqrt2\\=9\sqrt2.$
