## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$a_n=\dfrac{7}{15} \cdot 15^{n-1}$. and $a_n=-\dfrac{7}{15} \cdot (-15)^{n-1}$.
The $n^{th}$ term of the sequence is given by the formula: $a_n=a_1r^{n-1}$ where $r$=common ratio and $a_1$= the first term The common ratio of a geometric sequence is equal to the quotient of any term and the term before it: $\ r = \dfrac{a_n}{a_{n-1}}$ or, $r=\dfrac{a_2}{a_1}$ Here, we have: $a_4 = a_2 (r)(r); \\a_4 = a_2 \ ( r^2) \$ We will substitute the given values of $a_4$ and $a_2$ into the equation above to obtain: $1575 = 7 \times r^2 \implies \dfrac{1575}{7} = \dfrac{7\cdot r^2}{7} \implies 225 = r^2$ or, $r=\pm 15$ So, there are two possible values of $r$, $-15$ and $15$. Solve for the first term to obtain: Consider $r=15$, Now, $r=\dfrac{a_2}{a_1} \\15 = \dfrac{7}{a_1} \\ 15a_1 = 7 \\a_1 = \dfrac{7}{15}$ So, the $n^{th}$ term of the geometric sequence is: $a_n=\dfrac{7}{15} \cdot 15^{n-1}$. Next, consider $r=-15$, $r=\dfrac{a_2}{a_1} \\-15 = \dfrac{7}{a_1} \\-15a_1 = 7 \\a_1 = -\dfrac{7}{15}$ Therefore, the $n^{th}$ term of the geometric sequence is: $a_n=-\dfrac{7}{15} \cdot (-15)^{n-1}$.