How many terms are in each sequence?
12, 4, 4/3, ......, 4/729
help i dont know how to get answer
Hello, Vince604!
If you can't see the pattern, you need more help than we can offer.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 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.
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 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 $\displaystyle n.$