Question 173868
I am going to use the formula{{{A[n]=4(3^(n-1))}}} because the formula you gave makes no sense. 

so for the 12th term
:
{{{A[12]=4(3^(12-1))}}}
:
{{{A[12]=4(3^11)}}}
:
{{{A[12]=4(177147)}}}
:
{{{A[12]=highlight(708588)}}}