## Intermediate Algebra for College Students (7th Edition)

$a_7=8019$
Let us consider the sequence $a_n=a_13^{n-1}$ Now, $n=7$ and $a_1=11$ Thus, $a_7=a_13^{7-1}$ or, $a_7=a_13^{6}$ or, $a_7=(11)(729)=8019$ Hence, $a_7=8019$