Infinite Geometric Sequence: How to Find the Number of Terms

[Vince604]

Homework Statement

How many terms are in each sequence?

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

Homework Equations

The Attempt at a Solution

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

 

[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?

 

[Ibix]

Presuming you mean t_n=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?

 

[symbolipoint]

Big hint: <br /> 3^6 = 729<br />editted

 

[HallsofIvy]

Homework Statement

How many terms are in each sequence?

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

Homework Equations

The Attempt at a Solution

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

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?

 

[symbolipoint]

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

 

[Vince604]

Thanks