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Thread: arithmetic sequence problem

  1. #1
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    arithmetic sequence problem

    If "n+5, 2n+1, and 4n-3" are three consequtive terms in an arithmetic sequence, find n.
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  2. #2
    Super Member Anonymous1's Avatar
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    By inspection.

    $\displaystyle n=3$
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  3. #3
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    Hello, shawli!

    If $\displaystyle n+5,\; 2n+1,\; 4n-3$ are three consequtive terms
    in an arithmetic sequence, find $\displaystyle n.$

    $\displaystyle \text{We have: }\;\underbrace{n+5}_a,\:\underbrace{2n+1}_{a+d},\: \underbrace{4n-3}_{a+2d}$


    Then: .$\displaystyle \begin{array}{cccccccccc}n+5 + d &=& 2n+1 && \Rightarrow && n - d &=& 4 & [1] \\
    2n+1+d &=& 4n-3 && \Rightarrow && 2n-d &=& 4 & [2] \end{array}$


    Subtract [2] - [1]: .$\displaystyle \boxed{n \:=\:0}$

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  4. #4
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    Quote Originally Posted by shawli View Post
    If "n+5, 2n+1, and 4n-3" are three consequtive terms in an arithmetic sequence, find n.
    Hi shawli,

    Let's determine a common difference by subtracting the first term from the second term.

    $\displaystyle 2n+1-(n+5)=2n+1-n-5=\boxed{n-4} $

    Now determine the common difference between the 2nd and 3rd terms.

    $\displaystyle 4n-3-(2n+1)=4n-3-2n-1=\boxed{2n-4} $

    If it's truely an arithmetic sequence, these two common differences have to be equal.

    $\displaystyle 2n-4=n-4$

    Note to Anonymous1: I don't think 3 works.
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    Super Member Anonymous1's Avatar
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    Quote Originally Posted by masters View Post
    Note to Anonymous1: I don't think 3 works.
    Oh, I guess they have to be in the order given, then? Because $\displaystyle n=3$ yields the three consecutive integers $\displaystyle 7, 8, 9.$
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    Quote Originally Posted by Anonymous1 View Post
    Oh, I guess they have to be in the order given, then? Because $\displaystyle n=3$ yields the three consecutive integers $\displaystyle 7, 8, 9.$
    That was my assumption. Your order would've been 8, 7, 9 which wouldn't work.
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  7. #7
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, shawli!


    $\displaystyle \text{We have: }\;\underbrace{n+5}_a,\:\underbrace{2n+1}_{a+d},\: \underbrace{4n-3}_{a+2d}$


    Then: .$\displaystyle \begin{array}{cccccccccc}n+5 + d &=& 2n+1 && \Rightarrow && n - d &=& 4 & [1] \\
    2n+1+d &=& 4n-3 && \Rightarrow && 2n-d &=& 4 & [2] \end{array}$


    Subtract [2] - [1]: .$\displaystyle \boxed{n \:=\:0}$

    How can n=0??????????????
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  8. #8
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    Quote Originally Posted by Anonymous1 View Post
    How can n=0??????????????
    If you do what Soroban said, n is obviously equal to 0.

    You can review my post which is slightly different than Soroban's, but the conclusion is the same.

    n = 0

    Using that fact, the sequence becomes 5, 1, -3
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  9. #9
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    Ah , the answer actually IS n=0, but I came to the same conclusion as Anonymous1 and thought it didn't make sense.

    But now I get it ! Thanks!
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