Find the common ratio of the geometric series with the sum and the first term Given: Geometric Series Sum ($S_n$) = 39
First Term ($a_1$) = 3
number of terms ($n$) = 3
Find the common ratio $r$.
*I have been made aware that the the common ratio is 3, but for anyone trying to solve this, don't plug it in as an answer the prove that it's true. Try to find it without using it.
 A: If you have an infinite geometric series with first term $a$ and common ratio $r$ (with $|r|\lt1$) then the sum $s$ is $$s={a\over1-r}$$ 
In your problem, you know $s$ and $a$ --- can you solve for $r$?
EDIT: OP has clarified that the number of terms is $3$. 
That means the terms are $3$, $3r$, and $3r^2$. And they add up to $39$. So --- can you work out $r$ from this information?
A: You have $a+ar+ar^2=39$.  Do you understand how this comes from what you asked?  Then $a\frac  {r^3-1}{r-1}=39$ as shown in Wikipedia on geometric progression.  Given that $r=3$, we get $a\frac {26}2=39$ so $a=3$ and we have $3+9+27=39$.
A: Finite progression with $\,n\,$ elements:
$$39=S_n=3\frac{q^n-1}{q-1}\Longrightarrow \frac{q^n-1}{q-1}=13\ldots\text{we need to know}\,\,\,n$$
Infinite descending geometric series:
$$39=S=\frac{3}{1-q}\Longrightarrow 1-q=\frac{1}{13}\Longrightarrow q=\frac{12}{13}$$
A: You have the formula 
$$S_n=\frac{a*(1-r^n)}{1-r}$$
In your case, n=3, a=3 and $S_n$=39. Now solve for r.
A: The genreal sum formula in geometric sequences: $$S_n=\frac{a_1\left(r^n-1\right)}{r-1}$$ In your case, $S_n=39$, $a_1=3$, $n=3$. Now you just need to find $r$:
$$39=\frac{3\left(r^3-1\right)}{r-1}$$ $$\ 13=\frac{r^3-1}{r-1}$$ $$\ 13\left(r-1\right)=r^3-1$$ $$\ 13r-13=r^3-1$$ $$\ r^3-13r+12=0$$
$$\ r_1=-4,\ r_2=3,\ r_3=1$$
Since you said it's a geometric series, $r ≠ 1$, but even if it did, plugging it back in would give you $3 + 3 + 3 ≠ 39$. So, the common ratio is $$r = 3, -4$$ Plug $r=3$ back into the formula and see that it works. As for $r=-4$, you'll have to do it the long way because it won't work in the sum formula, so you could just write that $3 + (-12) + 48 = 39$ which is indeed true.
A: Any way one can find the ratio, r, for higher powers i.e. where n=12, a1 =80 and the sum is 1200.
