Given:
[tex]b(1)=-12[/tex]
[tex]b(n)=b(n-1)\cdot 4[/tex]
To find:
The third term of the sequence.
Solution:
We have,
[tex]b(n)=b(n-1)\cdot 4[/tex]
For n=2, we get
[tex]b(2)=b(2-1)\cdot 4[/tex]
[tex]b(2)=b(1)\cdot 4[/tex]
[tex]b(2)=-12\cdot 4[/tex] [tex][\because b(1)=-12][/tex]
[tex]b(2)=-48[/tex]
For n=3, we get
[tex]b(3)=b(3-1)\cdot 4[/tex]
[tex]b(3)=b(2)\cdot 4[/tex]
[tex]b(3)=-48\cdot 4[/tex] [tex][\because b(2)=-48][/tex]
[tex]b(3)=-192[/tex]
Therefore, the 3rd term of the sequence is -192.
