## College Algebra (10th Edition)

$a_n = 1 \cdot (-\frac{1}{3})^{n-1}$ $a_5=\frac{1}{81}$
RECALL: The $n^{th}$ term $a_n$ of a geometric sequence is given by the formula: $a_n=a_1 \cdot r^{n-1}$ where $a_1$ = first term $r$ = common ratio The given geometric sequence has: $a_1=1$ $r=-\frac{1}{3}$ Thus, the $n^{th}$ term of the sequence is given by the formula: $a_n = 1 \cdot (-\frac{1}{3})^{n-1}$ The 5th term can be found by substituting $5$ for $n$: $a_5=1 \cdot (-\frac{1}{3})^{5-1} \\a_5=1 \cdot (-\frac{1}{3})^4 \\a_5 = 1 \cdot \frac{1}{81} \\a_5=\frac{1}{81}$