Geometric Sequences, help please

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March 9th 2010, 08:38 AM
PiratePrincess

Geometric Sequences, help please

Hi there thanks for taking the time to read this :)
I'm studying for a test on sequences that I'll be taking this evening. I'm having trouble with questions like the one below. Can you please show me how it's suppose to be done?

In a geometric sequence, the tenth term is 2560 and the fifth term is 80. Find the twelfth term.

Thank you and have a great day!
March 9th 2010, 08:49 AM
masters

Quote:

Originally Posted by PiratePrincess

Hi there thanks for taking the time to read this :)
I'm studying for a test on sequences that I'll be taking this evening. I'm having trouble with questions like the one below. Can you please show me how it's suppose to be done?

In a geometric sequence, the tenth term is 2560 and the fifth term is 80. Find the twelfth term.

Thank you and have a great day!

Hi PiratePrincess,

To find any term in a geometric sequence, use

$a_n=a_1 \cdot r^{n-1}$

Between the 5th term and the 10th term including the 5th and 10th terms, there are 6 terms. So n = 6

$2560=80r^{6-1}$

$32=r^5$

$r=2$ "common ratio"

To find the 12th term, we can just start with the 10th term and multiply by the common ratio twice.

2560, 5120, 10240
March 9th 2010, 09:47 AM
PiratePrincess

Thank you so much (Clapping)

I wasn't too far off lol
I got to 32 = r^5 on my own but I don't understand the next step how did you get rid of the exponent?

Thanks again!
March 9th 2010, 09:52 AM
Soroban

Hello, PiratePrincess!

Masters did his usual masterful job.
Here's a slight different explanation . . .

Quote:

In a geometric sequence, the tenth term is 2560 and the fifth term is 80.
Find the twelfth term.

$\text{The }n^{th}\text{ term is: }\;a_n \:=\:ar^{n-1}\; \text{ where: }\:a = \text{first term, }\:r = \text{common ratio.}$

We are told: . $\begin{array}{ccccccc}a_{10} &=& ar^9 &=& 2560 & [1] \\ a_5 &=& ar^4 &=& 80 & [2] \end{array}$

Divide [1] by [2]: . $\frac{ar^9}{ar^4} \:=\:\frac{2560}{80}\quad\Rightarrow\quad r^5 \:=\:32 \quad\Rightarrow\quad r \:=\:2$

Substitute into [2]: . $a(2^4) \:=\:80 \quad\Rightarrow\quad 16a \:=\:80 \quad\Rightarrow\quad a \:=\:5$

Therefore: . $a_{12} \;=\;ar^{11} \;=\;5(2^{11}) \;=\;10,\!240$
March 9th 2010, 10:00 AM
masters

Quote:

Originally Posted by Soroban

Masters did his usual masterful job.
Here's a slight different explanation . . .

Thank you, Soroban. You are too kind.

Quote:

Originally Posted by PiratePrincess

Thank you so much (Clapping)

I wasn't too far off lol
I got to 32 = r^5 on my own but I don't understand the next step how did you get rid of the exponent?

Thanks again!

$r^5=32$

Take the 5th root of each side.

$\sqrt[5]{r^5}=\sqrt[5]{32}$

32 is a perfect 5th power.

$r=\sqrt[5]{2^5}$

$r=2$

Or, you could use rational exponents....

$r^5=2^5$

Therefore, $r=2$