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Arithemtic and geometric progession

  1. May 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Numbers a,b,c are consecutive members of increasing arithmetic progression, and numbers a,b,c+1 are consecutive members of geometric progression. If a+b+c=18 then a^2 +b^2 + c^2=?


    3. The attempt at a solution
    [itex]a + b + c= 18[/itex]

    [itex]a + a +d +a + 2d = 18[/itex]

    [itex]3a + 3d = 18[/itex]

    [itex]3(a+d)= 18[/itex]

    [itex]a+d=6=b[/itex]

    [itex]a + b + c + 1 = 19[/itex]

    [itex]a + aq + aq^{2} = 19[/itex]

    [itex]aq=b=6 \rightarrow q=\frac{6}{a} [/itex]

    [itex]a + aq + aq^{2} = 19 \rightarrow a +6 + \frac{36}{a} = 19 [/itex]

    [itex] a +6 + \frac{36}{a} = 19 /*a [/itex]

    [itex] a^{2} +6a + 36= 19a \rightarrow a^{2} -13a + 36= 0 [/itex]

    [itex]\frac{13 \pm \sqrt{169 - 144}}{2} = \frac{13 \pm 5}{2} [/itex]

    [itex]a_{1}=9[/itex] and [itex] a_{2}=4[/itex]

    If [itex] a_{1}=9[/itex] then [itex] d=-3 [/itex] but i have increasing arithemtic progression so i must take [itex] a_{2}=4[/itex].

    So [itex] a_{2}=4[/itex] and [itex] d=2 [/itex] since [itex]a+d=6[/itex].

    Then [itex] a=4[/itex], [itex] b=6[/itex], [itex] c=8[/itex]

    [itex] 4^{2} + 6^{2} + 8^{2} = 16 + 36 + 64 = 116 [/itex] but my textbooks says that [itex]133[/itex] is solution. What's the problem here?
     
    Last edited: May 28, 2013
  2. jcsd
  3. May 28, 2013 #2

    haruspex

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    Clearly your solution satisfies all the conditions. If you've quoted the problem correctly then it's an error in the book.
     
  4. May 28, 2013 #3
    the author might have accidentally plug in ( c + 1 ) as c in a^2+b^2+c^2...
    4^2 + 6^2 + 9^2 yields 133
     
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