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Math Help - 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.

    n=3
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  3. #3
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    Hello, shawli!

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

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


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


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

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  4. #4
    A riddle wrapped in an enigma
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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.

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

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

    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.

    2n-4=n-4

    Note to Anonymous1: I don't think 3 works.
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  5. #5
    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 n=3 yields the three consecutive integers 7, 8, 9.
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by Anonymous1 View Post
    Oh, I guess they have to be in the order given, then? Because n=3 yields the three consecutive integers 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!


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


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


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

    How can n=0??????????????
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  8. #8
    A riddle wrapped in an enigma
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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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