# Find the common ratio of the geometric series with the sum and the first term

Given:
Geometric Series Sum = 39
First Term = 3
number of terms = 3

*I do not know how to use special characters. I hope you can understand this. I need a formula for looking the common ratio of a geometric series. To make it more clear, the common ratio is 3.

• The solutions for this restricted problem are $r=3$ and $r=-4$. Just need to solve $1+r+r^2=13$. Sum of geometric formula is not useful here. – André Nicolas Feb 26 '13 at 5:17
• @AndréNicolas, I am very sure that the sum is correct and it should be included. (Sorry for unclear explanations) – clixbrigidxterx Feb 26 '13 at 5:25
• I meant that to solve this problem, you do not need the formula $a\frac{r^n-1}{r-1}$ for the sum of the GP $a+ar+\cdots +ar^{n-1}$. Since there are only $3$ terms, want $3+3r+3r^2=39$, which simplifies to $r^2+r-12=0$. A nice quadratic, it even factors. – André Nicolas Feb 26 '13 at 5:35
• 3,-12,48 works as does 3,9,27 so there are a pair of answers here. – JB King Feb 26 '13 at 5:38

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?

• Yes, I need the formula and how did you create that formula from the original geometric series formula. – clixbrigidxterx Feb 26 '13 at 5:05
• @clixbrigidxterx: If $a$ is the first term and $r$ is the ratio, you have $a(1+r+r^2)=39$. It seems likely that $a,r$ are integers, so factor $39$ and try them all. There aren't many. – Ross Millikan Feb 26 '13 at 5:08
• @clixbrigidxterx: if you know $r=3$ what is $1+r+r^2$? What do you know, it divides into $39$ – Ross Millikan Feb 26 '13 at 5:09
• @clixbrigidxterx Or you can compute the roots of the quadratic. – Julien Feb 26 '13 at 5:12
• @clixbrigidxterx did you need to construct the formula from scratch or do you just need to solve for r? I think Gerry explained it pretty well. Your first term is a, second term is the first term multiplied by r (ar), third term is the second multiplied by r ($ar^2$) etc. It's a good exercise to check that the formula from my answer holds for finite geometric series, and the formula Gerry wrote for his holds for infinite geometric series – user62228 Feb 26 '13 at 5:12

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 $3+9+27=39$.

• I want to see the process it gone through. Please, do not use the answer (common ratio) to substitute $r$. I only include this answer so you can prove that the answer is 3. – clixbrigidxterx Feb 26 '13 at 5:30
• @clixbrigidxterx: See André Nicolas' comment. He has it very simply for a 3 term progression. Then JB King solves the quadratic. For more general problems, you need to read on geometric progressions, which derive the sum of the series formula that we have all quoted. – Ross Millikan Feb 26 '13 at 14:04

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:

$$39=S=\frac{3}{1-q}\Longrightarrow 1-q=\frac{1}{13}\Longrightarrow q=\frac{12}{13}$$

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.

• @user6228, can you make show me how can I do it? I am stuck of finding the ratio in the formula because of exponents and some factors. – clixbrigidxterx Feb 26 '13 at 5:14