Chapter 12 Sequences and Series - 12.3 Analyze Geometric Sequences and Series - 12.3 Exercises - Skill Practice - Page 814: 15

$a_n=(-4)^{n-1}$ and $a_7=4096$

The nth term is given by $a_n= a_1 r^{n-1}$ Here, $a_1=1; r=\dfrac{a_2}{a_1}=\dfrac{-4}{1}=-4$ $a_n=(1) \times (-4)^{n-1}$ This gives: $a_n=(-4)^{n-1}$ Plug in $n=7$ $a_7=(-4)^{7-1}=4096$ Hence, $a_n=(-4)^{n-1}$ and $a_7=4096$