Question

The first term of a geometric series is -1, and the common ratio is -3. How many terms are in the series if its sum is 182?

Answer 1

<img src=" http://media.education2020.com/evresources/648-12-04-00-00_files/i0090000.jpg " alt="648-12-04-00-00_files/i0090000.jpg"/>

Answer 2

The geometric series is represented as:
\[a+ar+ar^2+...+ar^{n-1}=\sum_{k=0}^{n-1}ar^k=\sum_{k=1}^nar^{k-1} =a\frac{r^n-1}{r-1}\]
In your problem, \(a=-1\), and \(r=-3\) and it sums to 182
so:
\[ -1+(-1)(-3)+(-1)(-3)^2+...+(-1)(-3)^{n-1}\\=-1-(-3)-(-3)^2-...-(-3)^{n-1}\\ \, \\=-\frac{(-3)^n-1}{(-3)-1}
\\ \, \\
=-\frac{(-3)^n-1}{-4}=182\\
\, \\
=\frac{(-3)^n-1}{4}=182\\
(-3)^n-1=728\\
(-3)^n=729\\ n=6\]

there are 6 terms