## College Algebra (6th Edition)

Formula for the general term (the $n^{th}$ term) = $a_{1}$ + (n-1)d = 3n - 2
Given $2^{nd}$ term $a_{2}$ = $a_{1}$ + (2-1)d = $a_{1}$ + d = 4. $6^{th}$ term $a_{6}$ = $a_{1}$ + (6-1)d = $a_{1}$ + 5d = 16. where d = common difference in sequence On subtract $a_{2}$ from $a_{6}$. $a_{1}$ + 5d - $a_{1}$ - d = 16 - 4 4d = 12 d = 3 put the value of d in $a_{2}$ we got $a_{1}$ = 1. Now the sequence will be = 1, 4, 7, 10, 13, 16, . . . . Formula for the general term (the $n^{th}$ term) = $a_{1}$ + (n-1)d = 1 + (n - 1) 3 = 1 + 3n - 3 = 3n - 2