Chapter 12 Sequences and Series - 12.4 Find Sums of Infinite Geometric Series - 12.4 Exercises - Mixed Review - Page 825: 53

$a_n=6(-3)^{n-1}$

Here, we have $a_n= a_1r^{(n-1)}$ for the Geometric series. We are told that $a_2=-18$ and $r=-3$ Here, we have $a_2=(a_1) \times (-3)^{(2-1)} \implies -18=(a_1) \times (-3)$ or, $a_1=6$ Now, $a_n=a_1 \times r^{n-1}$ Hence, $a_n=6(-3)^{n-1}$