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Brut [27]
3 years ago
14

What is the sum of the geometric series in which a1 = 4, r = 3, and an = 324?

Mathematics
1 answer:
olganol [36]3 years ago
8 0
\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}\\\\

\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)

and surely you know how much that is.
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