## Precalculus (6th Edition) Blitzer

To ﬁnd the first four terms of the sequence whose general term is $a_{n}$ = 3n+2, we replace n in the formula with 1,2,3, and 4. n=1, $a_{1}$ = 3*1+2 = 3+2=5 n=2, $a_{2}$ = 3*2+2 = 6+2= 8 n=3, $a_{3}$ = 3*3+2 = 9+2 = 11 n=4,$a_{4}$ = 3*4+2 = 12+2 =14 The ﬁrst four terms are 5,8,11 and 14.