How many terms are in each sequence?

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

help i dont know how to get answer

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- Nov 15th 2012, 12:42 AMVince604geometric 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 AMa tutorRe: geometric sequence
Your terms are:

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

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 AMSorobanRe: 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?

The terms are successively divided by three.

So, the veryyou can do is*least***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 liketerms.*eight*

- Nov 15th 2012, 10:58 AMVince604Re: geometric sequence
- Nov 15th 2012, 11:30 AMSorobanRe: 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 n^{th}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 AMVince604Re: geometric sequence