Question

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

Answer 1

$\bf n^{th}\textit{ term of a geometric sequence}\\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=4\\ r=3\\ a_n=324 \end{cases} \implies 324=4(3)^{n-1} \\\\\\ \cfrac{324}{4}=3^{n-1}\implies 81=3^{n-1}\implies 3^4=3^{n-1}\implies 4=n-1 \\\\\\ \boxed{5=n}$

$\bf -------------------------------\\\\ \qquad \qquad \textit{sum of a finite geometric sequence}\\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=4\\ r=3\\ n=5 \end{cases} \\\\\\ S_5=4\left( \cfrac{1-3^5}{1-3} \right)\implies S_5=4\left(\cfrac{1-243}{-2} \right)$

and surely you know how much that is.