Answer
$601$ is the 33rd term of the sequence.
Work Step by Step
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.
