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

algebra-precalculusgeometric-progressionssequences-and-series

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?

Best Answer

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