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Math Help - Arithmetic series.

  1. #1
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    Arithmetic series.

    Not sure if this is the right section to be posting in but here goes.

    From a piece of wire 5000mm long, n pieces are cut, each being 10mm longer than the previous piece. If ui is the length of the ith piece and Si is the cumulative length of i pieces:

    (i) Calculate un and Sn in terms of u1 and n
    (ii) If exactly 25 pieces can be cut from the wire, determine the length of u1
    (iii) Determine the number of pieces that can be cut if u1 is 155mm


    Any pointers or help would be very welcome guys!
    Last edited by mr fantastic; December 20th 2010 at 01:09 PM. Reason: Re-titled.
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  2. #2
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    This is an arithmetic sequence problem, not a factor theorem one


    Note that there is a common difference between each length of wire

    i) In other words this is an arithmetic sequence with first time u1 and common difference 10: u_n = a+(n-1)d

    Similarly S_n = \dfrac{n}{2}(2a+(n-1)d)

    ii) find u(25) and solve for u_1

    iii) Solve for n using the sum formula given u_1 and d
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  3. #3
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    Quote Originally Posted by DanBrown100 View Post
    Not sure if this is the right section to be posting in but here goes.

    From a piece of wire 5000mm long, n pieces are cut, each being 10mm longer than the previous piece. If ui is the length of the ith piece and Si is the cumulative length of i pieces:

    (i) Calculate un and Sn in terms of u1 and n
    (ii) If exactly 25 pieces can be cut from the wire, determine the length of u1
    (iii) Determine the number of pieces that can be cut if u1 is 155mm


    Any pointers or help would be very welcome guys!
    Yes,

    this is an arithmetic sequence and series.

    (i)

    u_n=u_1+(n-1)10

    \displaystyle\ S_n=\frac{n}{2}\left[2u_1+(n-1)10\right]

    (ii)

    As S_n=5000

    5000=\displaystyle\frac{25}{2}\left[2u_1+240\right]

    from which the first term can be deduced.

    (iii)

    u_1=155\Rightarrow\frac{n}{2}\left[310+(n-1)10\right]=5000

    10000=300n+10n^2\Rightarrow\ n^2+30n-1000=0\Rightarrow\ (n+50)(n-20)=0
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