1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor

    "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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Number sequence
  1. Number Sequences (Replies: 10)

  2. Number Sequence (Replies: 1)

Loading...