College Algebra (10th Edition)

Two different geometric sequence satisfy the given, namely: $a_n=\frac{7}{15} \cdot (15)^{n-1}$ and $a_n=-\frac{7}{15} \cdot (-15)^{n-1}$
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 $a_2$ is known, the common ratio must be multiplied to $a_2$ to find the value of $a_4$. Thus, $a_4 = a_2 \cdot r \cdot r \\a_4 = a_2 \cdot r^2$ Substitute the given values of $a_4$ and $a_2$ into the equation above to obtain: $a_4 = a_2 \cdot r^2 \\1575 = 7 \cdot r^2 \\\dfrac{1575}{7} = \dfrac{7\cdot r^2}{7} \\225 = r^2$ Take the square root of both sides to obtain: $\pm \sqrt{225} = \sqrt{r^2} \\\pm 15 = r$ Thus, there are two possible values of $r$, $-15$ and $15$. Solve for the first term to obtain: If $r=15$, $r=\dfrac{a_2}{a_1} \\15 = \dfrac{7}{a_1} \\a_1(15) = \dfrac{7}{a_1} \cdot a_1 \\15a_1 = 7 \\a_1 = \dfrac{7}{15}$ Thus, the $n^{th}$ term of the geometric sequence is $a_n=\dfrac{7}{15} \cdot 15^{n-1}$. If $r=-15$, $r=\dfrac{a_2}{a_1} \\-15 = \dfrac{7}{a_1} \\a_1(-15) = \dfrac{7}{a_1} \cdot a_1 \\-15a_1 = 7 \\a_1 = \dfrac{7}{-15} \\a_1 = -\dfrac{7}{15}$ Thus, the $n^{th}$ term of the geometric sequence is $a_n=-\dfrac{7}{15} \cdot (-15)^{n-1}$.