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The sum of the series 6 + 18 +54 + ... to $n$ terms is $2184$. How many terms are in the series?

I know this series is geometric since it's constant by a common ratio.

I planned on using the formula: $S_n$$ = \frac{t_1(r^n-1)}{r-1}$

Where:

$S_n$ = sum of the first $n$ terms

$r$ = common ratio

$n$ = the number of terms

$t_1$ = first term

Here's what I did:

2184 = $\frac{6(3^n -1)}{2}$

2184 = $3(3^n -1)$

2184 = $9^n -3$

2187 = $9^n$

$ln(2187) = ln(9^n)$

$\frac{ln(2187)}{ln(9)} = \frac{n \times ln(9)}{ln(9)}$

$\frac{ln(2187)}{ln(9)} = n$

$3.5=n$

However, the answer in my text is $n=6$. I'm confused as to how I got this wrong?

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    $\begingroup$ $$3\cdot3^n \neq9^n$$ $\endgroup$ – John Glenn Jun 10 '18 at 5:04
  • $\begingroup$ @JohnGlenn would I just leave $9^n$ then, and distribute the 3 accordingly? $\endgroup$ – Jenny B Jun 10 '18 at 5:07
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$$3\cdot3^n\neq9^n\\ 3\cdot3^n=\color{red}{3^{n+1}}$$ But you're better off with: $$\frac{2184}{3}+1=3^n\\ 729=3^n\iff n=\frac{\ln729}{\ln3}=6$$

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