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Math Help - Geometric Sequences, help please

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by PiratePrincess View Post
    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
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  3. #3
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    Thank you so much

    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!
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  4. #4
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    Hello, PiratePrincess!

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


    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

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  5. #5
    A riddle wrapped in an enigma
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    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 View Post
    Thank you so much

    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
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