## College Algebra 7th Edition

$a=-384$ The $n^{th}$ term of the given sequence is: $a_n = -384 \cdot \left(-\dfrac{3}{8}\right)^{n-1}$.
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: $a_3=-54 \\a_6 =\dfrac{729}{256}$ Note that using the third term as reference point, the sixth term can be found by multiplying the common ratio $r$ three times to the value of the third term. Thus, $a_6=a_3 \cdot r \cdot r \cdot r \\a_6 = a_3 \cdot r^3$ Since $a_3=-54$ and $a_6=\dfrac{729}{256}$, substituting these values gives: $a_6=a_3 \cdot r^3 \\\dfrac{729}{256} = -54 \cdot r^3 \\\dfrac{1}{-54} \cdot \dfrac{729}{256} = \dfrac{1}{-54} \cdot (-54r^3) \\-\dfrac{27}{512}= r^3 \\(-\frac{3}{8})^3 = r^3$ Take the cube root of both sides to obtain: $-\dfrac{3}{8}=r$ With $r=-\dfrac{3}{8}$, the first term can be computed by dividing $a_3$ by the common ratio twice to obtain: $a = a_3 \div (r \times r) \\a = a_3 \div r^2 \\a=-54 \div (-\frac{3}{8})^2 \\a= -54 \div \frac{9}{64} \\a = -54 \cdot \frac{64}{9} \\a=-384$ 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: $a_n = -384 \cdot \left(-\dfrac{3}{8}\right)^{n-1}$