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Number sequence

  1. Jul 6, 2008 #1
    in a sequence, it is known that the multiplication of the 1st and second digits gives an answer or 4, the multiplication of the 3rd and 4th gives 70

    A1*A2=A1*(A1+d)=4
    A3*A4=(A1+2d)(A1+3d)=70

    A1^2+A1d=4

    A1^2+3A1d+2A1d+6d^2=70
    A1^2+5A1d+6d^2=70

    A1^2=4-A1d

    4-A1d+5A1d+6d^2=70
    4A1d+6d^2=66
    3d^2+2A1d=33

    from here i dont know what to do,
     
  2. jcsd
  3. Jul 6, 2008 #2

    uart

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    "from here i dont know what to do"

    Solve it as a quadratic to get a solution in the form of d as a function of A1. BTW this means there is not just one solution but a whole family of them.

    BTW. Is there any particular reason that you chose to use an arithmetic series? It's dead easy if you choose to use a geometric series.
     
  4. Jul 6, 2008 #3
    how do i do that?
     
  5. Jul 6, 2008 #4

    arildno

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    Incorrect.

    You have two equations in two unkwowns; you may reduce the system to one unknown being the solution of a fourth-order polynomial, yielding 4 solutions to the problem.
     
  6. Jul 6, 2008 #5

    uart

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    Do what, solve the quadratic or use a geometric series instead of an arithmetic series?

    To finish your current solution solve your last equation (3d^2+2A1d-33=0) using the quadratic formula to express d in terms of A1. Then continue (as arildno corrected me) and subtitute that back into the other equation (A1^2=4-A1d) to get an equation in terms of A1 alone. Solve this equation by re-arranging to get the sqare-root term on one side and all other terms on the other side of the equals. Now square both sides to get a 4th order polynomal which may require numerical methods to solve.

    Alternatively if there is no particular reason for using an arithmetic series instead of a gemetric series then just use the sequence a, ar, ar^2, ar^3 etc which makes the problem trivial to solve. (you get r^4 = 70/4 with very little effort).

    BTW. You still haven't told me why you feel that you should use an arithmetic series?
     
    Last edited: Jul 6, 2008
  7. Jul 6, 2008 #6

    arildno

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    As it happens, you'll get "lucky", in that the resulting fourth-order polynomial is really a second-order polynomial in the variable A^2. That can be readily solved, and if you are really good, your solutions will agree with mine:
    [tex]A_{1}^{(1)}=1,d^{(1)}=3[/tex]
    [tex]A_{1}^{(2)}=-1,d^{(2)}=-3[/tex]
    [tex]A_{1}^{(3)}=4\sqrt{3},d^{(3)}=-\frac{11}{\sqrt{3}}[/tex]
    [tex]A_{1}^{(4)}=-4\sqrt{3},d^{(4)}=\frac{11}{\sqrt{3}}[/tex]
     
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