# Thread: geometric sequence

1. ## geometric sequence

How many terms are in each sequence?

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

help i dont know how to get answer

2. ## Re: geometric sequence

$\displaystyle 12, \frac{12}{3}, \frac{12}{9}, \frac{12}{27}......\frac{12}{2187}$

Or

$\displaystyle \frac{12}{3^0}, \frac{12}{3^1}, \frac{12}{3^2}, \frac{12}{3^3}......\frac{12}{3^7}$

3. ## Re: geometric sequence

Hello, Vince604!

How many terms are in the sequence?

. . $\displaystyle 12,\;4,\;\tfrac{4}{3}\;\hdots\;\tfrac{4}{729}$

Help, I don't know how to get answer. . You don't?
If you can't see the pattern, you need more help than we can offer.

The terms are successively divided by three.

So, the very least you can do is list the terms.

. . $\displaystyle \begin{array}{ccccc}\text{start} && 12 & (1) \\ 12\div 3 &=& 4 & (2) \\ 4\div 3 &=& \frac{4}{3} & (3) \\ \\[-4mm] \frac{4}{3} \div 3 &=& \frac{4}{9} & (4) \\ \\[-4mm] \frac{4}{9} \div 3 &=& \frac{4}{27} & (5) \\ \\[-4mm] \frac{4}{27} \div 3 &=& \frac{4}{81} & (6) \\ \\[-4mm] \frac{4}{81}\div3 &=& \frac{4}{243} & (7) \\ \\[-4mm] \frac{4}{243}\div3 &=& \frac{4}{729} & (8) \end{array}$

Gee, it looks like eight terms.

4. ## Re: geometric sequence

Originally Posted by Soroban
Hello, Vince604!

If you can't see the pattern, you need more help than we can offer.

The terms are successively divided by three.

So, the very least you can do is list the terms.

. . $\displaystyle \begin{array}{ccccc}\text{start} && 12 & (1) \\ 12\div 3 &=& 4 & (2) \\ 4\div 3 &=& \frac{4}{3} & (3) \\ \\[-4mm] \frac{4}{3} \div 3 &=& \frac{4}{9} & (4) \\ \\[-4mm] \frac{4}{9} \div 3 &=& \frac{4}{27} & (5) \\ \\[-4mm] \frac{4}{27} \div 3 &=& \frac{4}{81} & (6) \\ \\[-4mm] \frac{4}{81}\div3 &=& \frac{4}{243} & (7) \\ \\[-4mm] \frac{4}{243}\div3 &=& \frac{4}{729} & (8) \end{array}$

Gee, it looks like eight terms.
Is there a faster way (like a formula i can use) to find the eighth term without having to list all the terms to get to it? Thanks for the help btw

5. ## Re: geometric sequence

Hello again, Vince604!

Is there a faster way (like a formula i can use) to find the eighth term,
without having to list all the terms to get to it?

Yes, there is a formula.

We have a geometric series.
The first term is $\displaystyle a = 12$; the common ratio is $\displaystyle r = \tfrac {1}{3}$

The nth term is: .$\displaystyle a_n \:=\:ar^{n-1} \:=\:12(\tfrac{1}{3})^{n-1}$

When does: .$\displaystyle 12\cdot\frac{1}{3^{n-1}} \:=\:\frac{4}{729}$ ?

Solve for $\displaystyle n.$

6. ## Re: geometric sequence

Originally Posted by Soroban
Hello again, Vince604!

Yes, there is a formula.

We have a geometric series.
The first term is $\displaystyle a = 12$; the common ratio is $\displaystyle r = \tfrac {1}{3}$

The nth term is: .$\displaystyle a_n \:=\:ar^{n-1} \:=\:12(\tfrac{1}{3})^{n-1}$

When does: .$\displaystyle 12\cdot\frac{1}{3^{n-1}} \:=\:\frac{4}{729}$ ?

Solve for $\displaystyle n.$
umm don't know