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Geometric Progression Problem

  1. Jul 14, 2008 #1
    Hi guys,
    In a geometric progression, the second term exceeds the first term by 20 and the fourth term exceeds the second term by 15. Find the possible values of the first term.

    ar - a = 20
    ar3 - ar = 15

    ar = 20 + a

    (20 + a)3 - (20 + a) = 15

    But this method seem to be wrong. Any help?
     
  2. jcsd
  3. Jul 14, 2008 #2

    HallsofIvy

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    Looks about as good as you could expect. it eleminates r. If you let x= 20+ a, then you have x3- x= 15. By the "rational root theorem", the only possible rational roots are integer divisors of 15: 1, 3, 5, and 15. It is easy to see that none of those work (13- 1= 0, 33- 3= 24, 53-5= 120, and 153- 15= 3360) and so x, and therefore a must be an irrational number. Probably a numerical solution is the best you could do.
     
  4. Jul 14, 2008 #3
    but the answer says -40 and -8. I don't get it @_@.
     
  5. Jul 14, 2008 #4

    HallsofIvy

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    Did you check that solution? if a= -40 and r= -8, then ar= 320 which is NOT 20 more than -40.

    Or if you meant a= -8 and r= -40, then ar= 320 again and the second terms is still NOT 20 more than the first term.

    Either the answer given is wrong or you are mis-reading the problem.
     
  6. Jul 14, 2008 #5
    it says, find the possible values of the first term. Values, so i assume, there is 2 possibility?
     
  7. Jul 14, 2008 #6

    rock.freak667

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    [tex]a(r-1)=20[/tex]

    [tex]ar(r^2-1)=15[/tex]



    try dividing them
     
  8. Jul 14, 2008 #7

    HallsofIvy

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    Ouch! I think I will go back and delete my post so I can pretend I never wrote it!
    That's very nice, rockfreak667.

    crays, your original mistake (and mine, since I didn't notice it) was thinking that since ar= 20+ a, ar3- ar= 15 would be (20+r)3- (20+r)= 15. That wrong because it is only r that is cubed in ar3. It is NOT
    (ar)3.
     
    Last edited: Jul 15, 2008
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