Answer:
The first six terms of the sequence are -3, 3, 21, 74, 234 and 714
Explanation:
Given the initial value a0 = -5 and the general form of the sequence given as An = 3an-1 + 12
To get the first six terms of the sequence, we will substitute for n=1 up to n= 6 in the general equation.
when n=1
A1 = 3a(1-1)+12
A1 = 3a0+12
A1 = 3(-5)+12
A1 = -15+12
A1 = -3
when n=2;

The first six terms of the sequence are -3, 3, 21, 74, 234 and 714