a_1&=-3
a_(n+1)&=a_n+8
To do so, we will use a table. To find a_2, we will substitute 1 for n in the above formula. To find a_3 we will substitute 2 for n, and so on.
n
a_(n+1)=a_n+8
a_n+8
a_(n+1)
-
a_1= -3
-
-
1
a_(1+1)=a_1+8 ⇕ a_2=a_1+8
a_2= a_1+8 ⇓ a_2= -3+8
a_2= 5
2
a_(2+1)=a_2+8 ⇕ a_3=a_2+8
a_3= a_2+8 ⇓ a_3= 5+8
a_3= 13
3
a_(3+1)=a_3+8 ⇕ a_4=a_3+8
a_4= a_3+8 ⇓ a_4= 13+8
a_4= 21
4
a_(4+1)=a_4+8 ⇕ a_5=a_4+8
a_5= a_4+8 ⇓ a_5= 21+8
a_5= 29
The first five terms of the sequence are -3, 5, 13, 21, and 29.
