## Intermediate Algebra (6th Edition)

$a_{n}= 42+8n$
Given sequence is $50,58,66....$ First term, $a_{1}= 50$ Common difference, $d = 58-50 = 8$ $a_{n}$ of the arithmetic sequence is $a_{n} = a_{1} + (n-1)d$ Substituting $a_{1}$ and $d$ $a_{n}= 50+(n-1)8$ $a_{n}=50+8n-8$ $a_{n}= 42+8n$