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    :dontknow::lol::lolz::question::thumbsup::zomg:Find four consecutive numbers in an A.P. such that when 2, 6, 7 and 2 are subtracted from these numbers respectively., the numbers are in geometric progression.


    Answer:5, 12, 19, 26

    How to solve it? Show the steps.:confused::bban::bban:
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    This looks like quite a fun question, I'll have a pop at it and get back to you.

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    (Original post by cucumberpj)
    Find four consecutive numbers in an A.P. such that when 2, 6, 7 and 2 are subtracted from these numbers respectively., the numbers are in geometric progression.


    Answer:5, 12, 19, 26
    As with the other question - can you show what you have done so far
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    (Original post by Krollo)
    This looks like quite a fun question, I'll have a pop at it and get back to you.

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    Thanks! I urgently need it.
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    (Original post by TenOfThem)
    As with the other question - can you show what you have done so far
    I really have no idea of it.
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    (Original post by cucumberpj)
    I really have no idea of it.
    Are you saying that you do not know what an AP or a GP is?
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    I haven't done it yet myself, but this is how I would start.

    You know the numbers are in AP, so you can represent them as a, a+d...

    Hint

    Following on, from the rules of GP, the ratio between each consecutive term must be the same.

    Therefore

    \dfrac{a+d-6}{a-2}=\dfrac{a+3d-7}{a+2d-2}
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    Let x, x+d, x+2d, and x+3d be the numbers in the arithmetic sequence.
    Let the numbers of the corresponding geometric sequence be:
    a, ar, ar^2, and ar^3.

    4 equation, 4 unknowns:
    x = a + 2 [1]
    x + d = ar + 6 [2]
    x + 2d = ar^2 + 7 [3]
    x + 3d = ar^3 + 2 [4]

    Substitute [1] into [2] and solve for d:
    a + 2 + d = ar + 6
    d = a(r - 1) + 4

    Substitute d and [1] into [3] and solve for a:
    a + 2 + 2[a(r - 1) + 4] = ar^2 + 7
    3 = ar^2 - 2ar + a
    3 = a(r - 1)^2
    3/(r - 1)^2 = a

    Substitute a, d, and [1] into [4] and solve for r:
    3/(r - 1)^2 + 2 + 3[(3/(r - 1)^2)(r - 1) + 4] =(3/(r - 1)^2)r^3 + 2
    3 + 12(r - 1)^2 + 9(r - 1) = 3r^3
    0 = 3(r^3 - 4r^2 + 5r - 2)
    0 = (r - 2)(r - 1)^2
    r = 1 or 2

    Substituting r = 1 into the expression for a won't work since we would be dividing by zero, so r = 2 is the only solution.

    a = 3/(2 - 1)^2 = 3

    d = 3(2 - 1) + 4 = 7

    x = 3 + 2 = 5

    x + d = 12
    x + 2d = 19
    x + 3d = 26


    any other simple method to solve it?
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    (Original post by cucumberpj)
    Let x, x+d, x+2d, and x+3d be the numbers in the arithmetic sequence.
    Use this and then the constant ratio idea that I showed you on the other thread

    You can do that twice here and then you have a simple quadratic
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    (Original post by TenOfThem)
    Use this and then the constant ratio idea that I showed you on the other thread

    You can do that twice here and then you have a simple quadratic
    x, x+d, x+2d, and x+3d
    =a , a+d, a+2d, a+3d

    G.P. : a-2 , a+d-6, a+2d-7, a+3d-2
    then apply common ratio, the equation has two unknowns, how to solve it?
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    (Original post by cucumberpj)
    x, x+d, x+2d, and x+3d
    =a , a+d, a+2d, a+3d

    G.P. : a-2 , a+d-6, a+2d-7, a+3d-2
    then apply common ratio, the equation has two unknowns, how to solve it?
    Well, you can do it twice

    \dfrac{a+3d-2}{a+2d-7} = \dfrac{a+2d-7}{a+d-6}

    and

    \dfrac{a+2d-7}{a+d-6} = \dfrac{a+d-6}{a-2}

    Two unknowns and two equations
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    (Original post by TenOfThem)
    Well, you can do it twice

    \dfrac{a+3d-2}{a+2d-7} = \dfrac{a+2d-7}{a+d-6}

    and

    \dfrac{a+2d-7}{a+d-6} = \dfrac{a+d-6}{a-2}

    Two unknowns and two equations
    thanks. I solved it. This method is easier to understand.
 
 
 
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