Question

How many terms are there in a geometric series if the first term is 4, the common ration is 3 and the sum of the series is 160

Answer 1

$\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=4\\ r=3\\ S_n=160 \end{cases}$

$\bf 160=4\left( \cfrac{1-3^n}{1-3} \right)\implies 160=4\left( \cfrac{1-3^n}{-2} \right)\implies 160=-2(1-3^n) \\\\\\ 160=2(3^n-1)\implies \cfrac{160}{2}=3^n-1\implies 80=3^n-1 \\\\\\ 81=3^n~~ \begin{cases} 81=3\cdot 3\cdot 3\cdot 3\\ \qquad 3^4 \end{cases}\implies 3^4=3^n\implies 4=n$