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Math Help - number of terms in geometric progression

  1. #1
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    number of terms in geometric progression

    obtain the number of terms in the following geometric progression

    81, 27 , 9 , 1/243

    can i find the number of terms using this

    ar^{n-1} = 1/243

    the answer that i get is a fraction though and that couldnt be possibly correct
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  2. #2
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    I suppose there are some skipped terms between 9 and 1/243.

    So a = 81 = 3^4 and r = 1/3. The equation is 3^4(1/3)^{n-1}=1/243=3^{-5}. From here, (1/3)^{n-1}=3^{-9}, or 3^{1-n}=3^{-9}, so n = 10.
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  3. #3
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    Hello, sigma1!

    \text{Obtain the number of terms in the following geometric progression:}

    . . . . 81,\;27,\;9,\;\hdots\;\dfrac{1}{243}

    This geometric progression has: . \begin{Bmatrix}\text{first term:} & a_1 &=& 81 \\ \text{common ratio:} & r &=& \frac{1}{3} \end{Bmatrix}

    \text{The }n^{th}\text{ term is: }\:a_n \:=\:a_1r^{n-1}

    So we have: . a_n \;=\; 81\left(\dfrac{1}{3}\right)^{n-1} \;=\;3^4\cdot\dfrac{1}{3^{n-1}} \;=\;\dfrac{1}{3^{n-5}}


    We want: . a_n \:=\:\dfrac{1}{243}

    . . \dfrac{1}{3^{n-5}} \:=\:\dfrac{1}{3^5} \quad\Rightarrow\quad n - 5 \:=\:5 \quad\Rightarrow\quad n \:=\:10


    Therefore, there are ten terms in the progression.

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  4. #4
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    Quote Originally Posted by emakarov View Post
    I suppose there are some skipped terms between 9 and 1/243.

    So a = 81 = 3^4 and r = 1/3. The equation is 3^4(1/3)^{n-1}=1/243=3^{-5}. From here, (1/3)^{n-1}=3^{-9}, or 3^{1-n}=3^{-9}, so n = 10.
    thanks alot.. i see where i went wrong.. if you were suposed to find the sum of the series though would you use this formula
    S_n = a(1-r^n) /1-r
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  5. #5
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    Quote Originally Posted by sigma1 View Post
    if you were suposed to find the sum of the series though would you use this formula
    S_n = a(1-r^n) /(1-r)
    Yes.
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