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.

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.
