Answer
$a_{70} =140\sqrt5$
Work Step by Step
RECALL:
(1) 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
(2) The common difference $d$ can be found by subtracting any term to the next term of the sequence:
$d=a_n - a_{n-1}$
The given sequence has:
$a_1=2\sqrt5$;
$d=4\sqrt5 - 2\sqrt5 = (4-2)\sqrt5=2\sqrt5$
Substitute these values into the formula for the $n^{th}$ term to obtain:
$a_n=a_1 + (n-1)d
\\a_n=2\sqrt5+(n-1)(2\sqrt5)$
To find the 80th term, substitute $80$ for $n$ to obtain:
$a_{70} = 2\sqrt5+(70-1)(2\sqrt5)
\\a_{70}=2\sqrt5+(69)(2\sqrt5)
\\a_{70}=2\sqrt5 + 138\sqrt5
\\a_{70} =140\sqrt5$
