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Math Help - Geometric Sequences and Series

  1. #1
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    Geometric Sequences and Series

    My Ďattemptí is pretty disastrous but I havenít found a better way to explain what Iím doing. Please bear with me.

    Q. The 1st, 3rd and 6th terms of an arithmetic sequence are also in geometric sequence. Find the common ratio (r) of the geometric sequence.

    A.:
    From text book, r = 3/ 2

    Attempt:
    If Un = a + (n - 1)d, assume d = 1, 2 and 3 respectively. This will result in the following sequences:

    If d = 1: Un-1 = 1, 2, 3, 4, 5, 6
    If d = 2: Un-1 = 2, 4, 6, 8, 10, 12
    If d = 3: Un-1 = 3, 6, 9 12, 15, 18

    From the question statement, we know that the 1st, 3rd and 6th numbers of the arithmetic sequence are in order for the geometric sequence.

    In sequence Un-1, where d = 1
    1, 3, 6 = U1, U2 and U3 of the geometric sequence respectively.

    Similarly, where d = 2, we have:
    2, 6, 12

    And when d = 3:
    3, 9, 18

    Note in each case: U3/ U2 = 2 and U2/ U1 = 3, e.g. in the sequence where d = 3:
    18/ 9 = 2 & 9/ 3 = 2

    Therefore, r = 3/2, i.e (U2/ U1)/ (U3/ U2)Ö?

    However,Ö
    Geometric sequence: Un = ar^n-1

    Going by geometric sequence layout, U1, U2, U3 = a, ar, ar^2, respectively.

    Which means that in, say Un-1, d = 3, the sequence 3, 9, 18 should really be 3, 9, 27, i.e we say r = 3 and so: U1 = 3, U2 = 3.3, U3 = 3.3^2

    Can anyone help me confirm if IĎm right, or completely off the mark? Thank you.
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  2. #2
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    Re: Geometric Sequences and Series

    Quote Originally Posted by GrigOrig99 View Post
    Q. The 1st, 3rd and 6th terms of an arithmetic sequence are also in geometric sequence. Find the common ratio (r) of the geometric sequence.
    A.:
    From text book, r = 3/ 2[B]
    You may be misreading this question.
    The first, third, and sixth terms are: a,~a+2d,~\&~a+5d.
    Those three terms form a geometric sequence so:
    ar=a+2d~\&~ar+2dr=a+5d.

    Can you solve those for r~?
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  3. #3
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    Re: Geometric Sequences and Series

    18/ 9 = 2 & 9/ 3 = 2

    Therefore, r = 3/2
    r is not (U3/U2)/(U2/U1),but it is U3/U2=U2/U1,and only when these two are equal.

    so none of the cases you mentioned satisfy the propositions.

    the answer:
    we have:
    U3=U2+d=(U1+d)+d=U1+2d
    and:
    U6=U5+d=...=(U1+4d)+d=U1+5d

    so (U1+5d)/(U1+2d)=(U1+2d)/U1
    because both are equal to r

    (U1+2d)(U1+2d)=U1(U1+5d)
    U1^2+4dU1+4d^2=U1^2+5dU1
    subtract U1 from both sides:
    4dU1+4d^2=5dU1
    4d^2=dU1
    divide both by d
    4d=U1

    next:
    r=U3/U1
    r=(U1+2d)/U1
    r=(4d+2d)/4d
    r=6d/4d
    r=6/4
    r=3/2

    hope it helps!
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    Re: Geometric Sequences and Series

    Ok, I'm starting to understand this but, Anonimnystefy, at the end of your explanation...

    next:
    r = U3/U1
    How did you arrive at U3/ U1? Is it not meant to be U2/U1?

    And Plato, when you say:
    How did you get ar + 2dr = a +5d? Is it not ar^2 = a+ 5d?

    I don't mean to come across like an ass, but if I understand how you worked those parts out, I should be able to piece this together. Thank you for all the help so far.
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  5. #5
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    Re: Geometric Sequences and Series

    Quote Originally Posted by GrigOrig99 View Post
    And Plato, when you say:
    How did you get ar + 2dr = a +5d? Is it not ar^2 = a+ 5d?
    In any geometric sequence each term is the same multiple of the preceding term.
    So if g_1,~g_2,~\&~g_3 is a geometric sequence then g_2=g_1r,~\&~g_3=g_2r=g_1r^2.
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  6. #6
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    Re: Geometric Sequences and Series

    Ok, so can I say:

    ar = a + 2d => r = (a + 2d)/ a => r = 2d

    ar^2 = a + 5d => r(ar) = a + 5d => r(a + 2d) = a + 5d => ar + 2dr = a + 5d => r(a + 2d) = a + 5d => r = (a + 5d)/ (a + 2d) => r = 3d

    r = 3d/ 2d => r = 3/2
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  7. #7
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    Re: Geometric Sequences and Series

    Yes. That will work.
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  8. #8
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    Re: Geometric Sequences and Series

    Thank you very much. I genuinely appreciate the help.
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