## College Algebra (10th Edition)

$a_{70} =140\sqrt5$
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: $a_{70} = 2\sqrt5+(70-1)(2\sqrt5) \\a_{70}=2\sqrt5+(69)(2\sqrt5) \\a_{70}=2\sqrt5 + 138\sqrt5 \\a_{70} =140\sqrt5$