Answer
$a_n=\dfrac{7}{15} \cdot 15^{n-1}$.
and $a_n=-\dfrac{7}{15} \cdot (-15)^{n-1}$.
Work Step by Step
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: $a_n=-\dfrac{7}{15} \cdot (-15)^{n-1}$.
