Chapter 11 - 11.3 - Geometric Sequences and Series - 11.3 Exercises - Page 795: 41

$a_9=\frac{1}{768}$

$a_1=\frac{1}{3}$ $r=\frac{a_2}{a_1}=\frac{a_3}{a_2}$ $r=\frac{-\frac{1}{6}}{\frac{1}{3}}=\frac{\frac{1}{12}}{-\frac{1}{6}}=-\frac{1}{2}$ $a_n=a_1r^{n-1}$ $a_9=\frac{1}{3}(-\frac{1}{2})^{9-1}=\frac{1}{3}(-\frac{1}{2})^8=\frac{1}{3}(\frac{1}{256})=\frac{1}{768}$