## Precalculus (6th Edition) Blitzer

$8019$
The sum of an arithmetic sequence is given by: $S_n=\dfrac{n}{2}[a_1+a_n]$ and the nth term for an arithmetic sequence is given by $a_n=a_1+(n-1) d$ We are given that $11,33, 99, 297,......$ This gives: $a_n=a_1 \cdot 3^{n-1}$ Here, $a_7=11 \cdot 3^{7-1} =11 \cdot 3^{6}$ Hence, $a_7=8019$