## College Algebra 7th Edition

$601$ is the 33rd term of the sequence.
RECALL: The $n^{th}$ $a_n$ of an arithmetic sequence is given by the formula: $a_n = a + d(n-1)$ where $a$ = first term $d$ = common difference The sequence has $a=25$ and $d=18$. Substituting these tot he formula above gives: $a_n=25 + 18(n-1)$ The $n^{th}$ term is $601$. To know the value of $n$, substitute $601$ to $a_n$ to obtain: $a_n = 25+18(n-1) \\601=25+18(n-1) \\601 - 25 = 18(n-1) \\576 = 18(n-1) \\\dfrac{576}{18}=\dfrac{18(n-1)}{18} \\32=n-1 \\32+1 = n-1+1 \\33=n$ Thus, $601$ is the 33rd term of the sequence.