## College Algebra 7th Edition

$a = \dfrac{81}{2}$ The $n^{th}$ term of the given sequence is: $a_n = \dfrac{81}{2} \cdot \left(\dfrac{2}{3}\right)^{n-1}$
To find the first and nth terms, the value of the common ratio $r$ is needed. Note that the next term of a geometric sequence can be found by multiplying the common ratio $r$ to the current term. The known terms are: $a_4=12 \\a_7 =\dfrac{32}{9}$ Note that using the fourth term as reference point, the seventh term can be found by multiplying the common ratio $r$ three times to the value of the fourth term. Thus, $a_7=a_4 \cdot r \cdot r \cdot r \\a_7 = a_4 \cdot r^3$ Since $a_4=12$ and $a_7=\dfrac{32}{9}$, substituting these values gives: $a_7=a_4 \cdot r^3 \\\dfrac{32}{9} = 12 \cdot r^3 \\\dfrac{1}{12} \cdot \dfrac{32}{9} = \dfrac{1}{12} \cdot 12r^3 \\\dfrac{8}{27} = r^3 \\(\frac{2}{3})^3 = r^3$ Take the cube root of both sides to obtain: $\dfrac{2}{3}=r$ With $r=\dfrac{2}{3}$, the first term can be computed by dividing $a_4$ by the common ratio three times to obtain: $a = a_4 \div (r \times r \times r) \\a = a_4 \div r^3 \\a=12 \div (\frac{2}{3})^3 \\a= 12 \div \dfrac{8}{27} \\a = 12 \cdot \dfrac{27}{8} \\a = \dfrac{81}{2}$ RECALL: The $n^{th}$ term $a_n$ of a geometric sequence can be found using the formula $a_n = a \cdot r^{n-1}$ where $a$ = first term $r$ = common ratio Thus, the $n^{th}$ term of the given sequence is: $a_n = \dfrac{81}{2} \cdot \left(\dfrac{2}{3}\right)^{n-1}$