Answer
$a_n=\sqrt2+(n-1)(\sqrt2)$
$a_{51}=51\sqrt2$
Work Step by Step
RECALL:
The $n^{th}$ term of an arithmetic sequence can be found using the formula:
$a_n = a_1 + (n-1)d$
where $a_1$ = first term and $d$ = common difference
The given sequence has:
$a_1=\sqrt2$;
$d=\sqrt2$
Substitute these values into the formula for the $n^{th}$ term to obtain:
$a_n=a_1 + (n-1)d
\\a_n=\sqrt2+(n-1)(\sqrt2)$
To find the 51st term, substitute $51$ for $n$ to obtain:
$a_{51} = \sqrt2+(51-1)\sqrt2
\\a_{51}=\sqrt2+50\sqrt2
\\a_{51}=51\sqrt2$
