Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Skill Practice - Page 814: 22

$a_n=-0.3(-2)^{n-1}$ and $a_7=-19.2$

The nth term is given by $a_n= a_1 r^{n-1}$ ...(1) Here, $a_1=-0.3; r=\dfrac{a_2}{a_1}=\dfrac{0.6}{-0.3}=-2$ From equation (1), we have $a_n=-0.3(-2)^{n-1}$ Plug in $n=7$ $a_7=-0.3(-2)^{7-1}=-19.2$ Hence, $a_n=-0.3(-2)^{n-1}$ and $a_7=-19.2$