Respuesta :
[tex]\bf 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}\\
----------\\
a_1=4\\
r=3\\
a_n=324
\end{cases}
\implies
324=4(3)^{n-1}
\\\\\\
\cfrac{324}{4}=3^{n-1}\implies 81=3^{n-1}\implies 3^4=3^{n-1}\implies 4=n-1
\\\\\\
\boxed{5=n}\\\\[/tex]
[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}\\ ----------\\ a_1=4\\ r=3\\ n=5 \end{cases} \\\\\\ S_5=4\left( \cfrac{1-3^5}{1-3} \right)\implies S_5=4\left(\cfrac{1-243}{-2} \right)[/tex]
and surely you know how much that is.
[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}\\ ----------\\ a_1=4\\ r=3\\ n=5 \end{cases} \\\\\\ S_5=4\left( \cfrac{1-3^5}{1-3} \right)\implies S_5=4\left(\cfrac{1-243}{-2} \right)[/tex]
and surely you know how much that is.
