so, we know that r = -3, and that the sum of the first 5 terms is 427 hmmmm
[tex]\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\
S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
n=5\\
r=-3\\
S_5=427
\end{cases}
\\\\\\
S_5=a_1\left( \cfrac{1-r^5}{1-r} \right)\implies 427=a_1\left( \cfrac{1-(-3)^5}{1-(-3)} \right)[/tex]
[tex]\bf 427=a_1\left(\cfrac{1+243}{1+3} \right)\implies 427=a_1(61)\implies \cfrac{427}{61}=a_1\implies 7=a_1
\\\\\\
n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
r=-3\\
n=5\\
a_1=7
\end{cases}
\\\\\\
a_5=7 (-3)^{5-1}\implies a_5=7(-3)^4[/tex]
and surely you know how much that is.