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Math Help - there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1thenx1=?

  1. #1
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    there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1thenx1=?

    More precisely,
    Let x1,x2,....x100 be positive nos. such that the sum of two adjacent nos. is k(constant). (xi +x (i+1) = k for all i)
    If x10=1
    x1=?

    I solved it and my answer is one. But at the back it is showing k-1.
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  2. #2
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    Re: there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1then

    Quote Originally Posted by AaPa View Post
    More precisely,
    Let x1,x2,....x100 be positive nos. such that the sum of two adjacent nos. is k(constant). (xi +x (i+1) = k for all i)
    If x10=1
    x1=?

    I solved it and my answer is one. But at the back it is showing k-1.
    $x_{2n}=1$

    $x_{2n+1}=k-1$

    1 is odd so $x_1=k-1$
    Thanks from AaPa
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  3. #3
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    Re: there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1then

    nice method. Thanks. I had done it with another method but wrongly. I corrected it-right answer.
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  4. #4
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    Re: there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1then

    Hi,
    When I was teaching I would discuss the sequence described in the attachment in many different classes. The main purpose was to show there is nothing mysterious about compound interest, examples 2 and 3.

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  5. #5
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    Re: there are 100 nos. x1,x2,..such that sum of 2 adjacent nos is k(const.).x10=1then

    cool
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