[tex]\qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle 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=\textit{last term's}\\ \qquad position\\ a_1=\stackrel{\textit{first term}}{-3}\\ r=\stackrel{\textit{common ratio}}{6} \end{cases}[/tex]
[tex]\stackrel{S_n}{-4665}~~ = ~~-3\left( \cfrac{1-6^n}{1-6} \right)\implies -4665=-3\left( \cfrac{1-6^n}{-5} \right) \\\\\\ -4665=\cfrac{3}{5}(1-6^n)\implies -4665\cdot \cfrac{5}{3}=1-6^n\implies -7775=1-6^n \\\\\\ -7776=-6^n\implies 7776=6^n\implies 6^5=6^n\implies \boxed{5=n}[/tex]
Answer:
5 terms
Step-by-step explanation:
The attached table shows that the sum of 5 terms is -4665.
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Additional comment
A table for a problem like this is easily made using a spreadsheet. Each value is 6 times the previous, and each sum is the new value added to the previous sum. The first value and the first sum are -3.
In the attached, we have simply summed the terms, each term being computed using the explicit formula for this geometric sequence.
