# geometric sequence

• Nov 15th 2012, 12:42 AM
Vince604
geometric sequence
How many terms are in each sequence?

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

help i dont know how to get answer
• Nov 15th 2012, 01:50 AM
a tutor
Re: geometric sequence

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

Or

$\displaystyle \frac{12}{3^0}, \frac{12}{3^1}, \frac{12}{3^2}, \frac{12}{3^3}......\frac{12}{3^7}$
• Nov 15th 2012, 05:53 AM
Soroban
Re: geometric sequence
Hello, Vince604!

Quote:

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.
• Nov 15th 2012, 10:58 AM
Vince604
Re: geometric sequence
Quote:

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
• Nov 15th 2012, 11:30 AM
Soroban
Re: geometric sequence
Hello again, Vince604!

Quote:

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 $\displaystyle n.$
• Nov 15th 2012, 11:53 AM
Vince604
Re: geometric sequence
Quote:

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