## College Algebra (10th Edition)

$a_n = 3 \cdot (\frac{1}{3})^{n-1}$
RECALL: (1) 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 (2) To find the next term in a geometric sequence, the common ratio $r$ is multiplied by the current term. Part (2) above implies that if the value of $a_3$ is known, the common ratio must be multiplied by $a_3$ three times to find the value of $a_6$. Thus, $a_6 = a_3 \cdot r \cdot r \cdot r \\a_6 = a_3 \cdot r^3$ Substitute the given values of $a_3$ and $a_6$ into the equation above to obtain: $a_6 = a_3 \cdot r^3 \\\frac{1}{81} = \frac{1}{3} \cdot r^3$ Multiply by $3$ on both sides to obtain: $3 \cdot \frac{1}{81} = \frac{1}{3} \cdot r^3 \cdot 3 \\\frac{1}{27} = r^3 \\(\frac{1}{3})^3=r^3$ Take the cube root of both sides to obtain: $\sqrt[3]{(\frac{1}{3})^3} = \sqrt[3]{r^3} \\\frac{1}{3} = r$ Note that: $a_3=a_1 \cdot r \cdot r \\a_3 = a_1 \cdot r^2$ Substitute the known values of $a_3$ and $r$ into the equation above to obtain: $a_3 = a_1 \cdot r^2 \\\frac{1}{3} = a_1 \cdot (\frac{1}{3})^2 \\\frac{1}{3} = a_1 \cdot \frac{1}{9}$ Multiply by $9$ on both sides of the equation to obtain: $9(\frac{1}{3}) = a_1 \cdot \frac{1}{9} \cdot 9 \\3 = a_1$ Thus, the $n^{th}$ term of the geometric sequence is: $a_n=a_1 \cdot r^{n-1} \\a_n = 3 \cdot (\frac{1}{3})^{n-1}$