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Arithmetic and Geometric sequence problem

  1. Dec 5, 2015 #1
    1. The problem statement, all variables and given/known data
    The sum of first three numbers of the arithmetic sequence is 54. If you subtract 3 from the first one, leave the second one unchanged and add 12 to the third one you get the first three numbers of the geometric sequence of the form ##ar + ar^2 + ar^3 + ... ar^n ## Find r.

    2. Relevant equations
    3. The attempt at a solution

    Using the first clue i got ##a + d = 18 ## ##a ## being the first number in either sequence and ##d ## being the difference of the arithmetic sequence. Next i set up the geometric sequence of the given form ##(a - 3) + (a + d) + (a + 2d + 12) ## . Now ##\frac{a + d}{a - 3} = r ## and from this ##\frac{18}{18 - d} = r ## . Also ##\frac{a + 2d + 12}{a + d} = r^2 ## . Now defining ##18 - \frac{18}{r} = d ## and replacing in the formula above i obtain ##8r - 3 = r^3## . This does not hold for ##r = 2 ## which is the correct solution. Which step did i do wrong?
     
    Last edited by a moderator: Dec 5, 2015
  2. jcsd
  3. Dec 5, 2015 #2

    Samy_A

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    When you went from ##\frac{a + d}{a - 3} = r## to ##\frac{18}{18 - d} = r##, you forgot the ##-3## of the first denominator.
     
  4. Dec 5, 2015 #3
    True but didnt make much difference. i now get ##5r - 2 = 2r^3 ## Cant find any mistakes but there must be some.
     
  5. Dec 5, 2015 #4

    Samy_A

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    ##\frac{a + 2d + 12}{a + d} = r^2 ## is not correct.
     
    Last edited: Dec 5, 2015
  6. Dec 5, 2015 #5
    yeah, thats it. Just r there. Now its correct ##5r - 2 = 2r^2 ##
     
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