Answer
Two different geometric sequence satisfy the given, namely:
$a_n=\frac{7}{15} \cdot (15)^{n-1}$
and
$a_n=-\frac{7}{15} \cdot (-15)^{n-1}$
Work Step by Step
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 $a_n=-\dfrac{7}{15} \cdot (-15)^{n-1}$.
