## Precalculus (10th Edition)

$a_n=n\sqrt2.$ Therefore $a_{51}=51\sqrt2.$
We know that $a_n=a_1+(n-1)d.$ where $d$=common difference and $a_1$= the first term Hence, here we have $a_n=\sqrt2+(n-1)\cdot(\sqrt2)\\=n\sqrt2.$ Therefore $a_{51}=51\sqrt2.$