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Math Help - Sequence and series question

  1. #1
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    Sequence and series question

    I am given the common difference/ratio and tn = 48 and Sn = 33, I need to find the first term but I am not given any terms how do I do this?
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  2. #2
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    Re: Sequence and series question

    The common difference or the common ratio? I would need to know to determine if this is arithmetic or geometric!
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  3. #3
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    Re: Sequence and series question

    Hello, Cryptnar!

    I've made some progress . . .


    \text{I am given the common difference/ratio and }t_n = 48\text{ and }S_n = 33.

    \text{I need to find the first term, but I am not given any terms. \; How do I do this?}

    Suppose we have a Geometric Progression.

    We are given: . \begin{Bmatrix}t_n &=& ar^{n-1} &=& 48 & [1] \\ \\[-3mm] S_n &=& a\dfrac{1-r^n}{1-r} &=& 33 & [2] \end{Bmatrix}


    From [2]: . a(1-r^n) \:=\:33(1-r) \quad\Rightarrow\quad a - ar^n \:=\:33 - 33r\;[3]


    Multiply [1] by r\!:\;ar^n \:=\:48r


    Substitute into [3]: . a - 48r \:=\:33 - 33r


    . . Therefore: . a \:=\:33 + 15r\;\hdots\;\text{ and we are given }r.
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  4. #4
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    Re: Sequence and series question

    Quote Originally Posted by Soroban View Post
    Hello, Cryptnar!

    I've made some progress . . .



    Suppose we have a Geometric Progression.

    We are given: . \begin{Bmatrix}t_n &=& ar^{n-1} &=& 48 & [1] \\ \\[-3mm] S_n &=& a\dfrac{1-r^n}{1-r} &=& 33 & [2] \end{Bmatrix}


    From [2]: . a(1-r^n) \:=\:33(1-r) \quad\Rightarrow\quad a - ar^n \:=\:33 - 33r\;[3]


    Multiply [1] by r\!:\;ar^n \:=\:48r


    Substitute into [3]: . a - 48r \:=\:33 - 33r


    . . Therefore: . a \:=\:33 + 15r\;\hdots\;\text{ and we are given }r.
    Hey, I don't really understand the a-arn thing what does the "a-" represent?
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  5. #5
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    Re: Sequence and series question

    It's the first term in the sequence
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