# Geometric Sequence i'm lost

1. Nov 15, 2012

### Vince604

1. The problem statement, all variables and given/known data
How many terms are in each sequence?

12, 4, 4/3, ......, 4/729

2. Relevant equations

3. The attempt at a solution

using tn=t1(r)(n-1) ??? im lost

2. Nov 15, 2012

### symbolipoint

That sequence clearly shows each n+1 term is the n term multiplied by $\frac{1}{3}$

Even the denominator of the last term is a multiple of 3 and only 3 (and 1, which is redundant); while the very first term, 12, is a whole positive integer.

Denominator of the n=3 term is 3^1. Denominator of the last term is 3^p. What is p? How does this relate the the value of n for this last term?

Last edited: Nov 15, 2012
3. Nov 15, 2012

### Ibix

Presuming you mean tn=t1r(n-1), then you are doing fine. What do each of the terms mean? Which ones do you know and which one are you supposed to evaluate? What are you missing and how do you figure it out?

4. Nov 15, 2012

### symbolipoint

Big hint: $3^6 = 729$

editted

Last edited: Nov 15, 2012
5. Nov 15, 2012

### HallsofIvy

Staff Emeritus
This should be t_1 r^(n-1).

t_1 is the first number, right? And that is 12. What is r? To go from 12 to 4, you divide by 3, to go from 4 to 4/3, you divide by 3.... Add dividing by 3 is the same as multiplying by 1/3. So t_n= (12)(1/3)^(n-1)= 4/729. Divide on both sides by 12: t_n= (1/3)^(n-1)= 1/2187. That is the same as 3^(n-1)= 2187. You could answer that by taking logarithms but it is perhaps simper to do as symbolipoint suggested: look at powers of 3. 3^2= 9, 3^3= 27, 3^4= 81, 3^5= 243, 3^5= 729, 3^6= 2187. What is n-1? And so what is n?

6. Nov 15, 2012

### symbolipoint

HallsofIvy,
At least I had the right idea, but obviously I made an arithmetic mistake. 3^5 vs. 3^6

7. Nov 15, 2012

Thanks