a_1&=-4
a_(n+1)&=2a_n-6
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)=2a_n-6
2a_n-6
a_(n+1)
-
a_1= -4
-
-
1
a_(1+1)=2a_1-6 ⇕ a_2=2a_1-6
a_2=2 a_1-6 ⇓ a_2=2( -4)-6
a_2= -14
2
a_(2+1)=2a_2-6 ⇕ a_3=2a_2-6
a_3=2 a_2-6 ⇓ a_3=2( -14)-6
a_3= -34
3
a_(3+1)=2a_3-6 ⇕ a_4=2a_3-6
a_4=2 a_3-6 ⇓ a_4=2( -34)-6
a_4= -74
4
a_(4+1)=2a_4-6 ⇕ a_5=2a_4-6
a_5=2 a_4-6 ⇓ a_5=2( -74)-6
a_5= -154
The first five terms of the sequence are -4, -14, -34, -74, and -154.
