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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






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


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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


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    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


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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


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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:
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