Chapter 9 - Conic Sections, Sequences, and Series - 9.4 Geometric Sequences - 9.4 Exercises - Page 738: 35

$a_n=3\left(-2\right)^{n-1}$

$a_4=-24$ and $a_{11}=3072$ $\frac{a_{11}}{a_4}=\frac{ar^{10}}{ar^3}=\frac{3072}{-24}=-128=r^7$ $\implies r=\left(-128\right)^{\frac{1}{7}}=-2.$ $a_4=ar^3=a\left(-2\right)^3=a\left(-8\right)=-24$ $\implies a=3$ The $n^{th}$ term is therefore $a_n=3\left(-2\right)^{n-1}$