Finding the Number of Terms (n) in a Geometric Series When Given $t_1$ to $t_3$ and The Sum? 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?

$$3\cdot3^n\neq9^n\\
3\cdot3^n=\color{red}{3^{n+1}}$$
But you're better off with:
$$\frac{2184}{3}+1=3^n\\
729=3^n\iff n=\frac{\ln729}{\ln3}=6$$
