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Homework Help: Geometric Sequences and Series

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Q.: Show that if log a, log b and log c are three consecutive terms of an arithmetic sequence, then a, b and c are in geomtric sequence.

    2. Relevant equations

    Un = a + (n - 1)d and Sn = [itex]\frac{a(r^n - 1)}{r - 1}[/itex]

    3. The attempt at a solution

    Consider arithmetic sequence log 2, log 4, log 8 or log 3, log 9, log 27

    The respective values of each sequence is 0.301, 0.602, 0.903 and 0.477, 0.954, 1.431

    The difference between the numbers in the first sequence is 0.301 and 0.477 in the second.

    Geoetric sequence is a, ar, ar^2. So it follows from the above sequnces that...
    a = 2 or 3
    ar = 2.2 = 4 or 3.3 = 9
    ar^2 = 2.2^2 = 8 or 3.3^2 = 27

    Thus a, b and c can be expressed in arithmetic sequence as log values and then as geometric sequences without log.

    Answer: From text book: b^2 = ac [itex]\Rightarrow[/itex] Geometric sequence.

    I feel I've adequately proven the point, albeit in a slightly unconventional way. Can someone steer me toward the book's answer, or is my method already satisfactory? Thank you.
  2. jcsd
  3. Aug 22, 2011 #2


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    well if loga,logb,logc are the terms then what is d? (equate 'd' for the first two and 'd' for the second two).

    What will you get then?
  4. Aug 22, 2011 #3


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    write the sequence as: [itex]\log a=\alpha +(n-1)d[/itex], [itex]\log b=\alpha +nd[/itex] and [itex]\log c=\alpha +(n+1)d[/itex]. You should be able to compute d in two ways, this will give you a relationship between a,b and c. You do the same sort of thing with a,b and c as a geometric series.
  5. Aug 22, 2011 #4


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    Hmm. How about taking the recursive definition of a geometric series and taking the log of it. Seems completely trivial.

    [Edit: sequence, not series; Note the OP problem statement; that is what I meant].
    Last edited: Aug 22, 2011
  6. Aug 22, 2011 #5
    Here's the second shot...

    In an arithmetic sequence U2 - U1 = d
    Therefore... (log b = a + nd) - (log a = a + nd - d)
    b - a = d

    In a geometric sequence U2/ U1 = r
    Therefore... (log b = [itex]\frac{a(r^n - 1)}{r - 1}[/itex]) / (log a = [itex]\frac{a(r^n - 1}{r - 1}[/itex])

    b/ a = [itex]\frac{ar^n(r - 1)}{r(ar^n - a)}[/itex]

    b/ a = [itex]\frac{ar^n.r - ar^n}{ar^n.r - ar}[/itex]

    b/ a = [itex]\frac{1 - ar^n}{1 - ar}[/itex]

    Thats what I have at the moment. The revised arithmetic sequence looks correct but I'm not sure about the geometric sequence. Can someone verify if I'm on track? Thank you.
  7. Aug 22, 2011 #6


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    If your three terms in AP are loga,logb,logc; using for formula for d then

    d = u2-u1= logb-loga.

    Shouldn't d also be equal to u3-u2 ?

    Similarly for the GP, you are correct in the r = b/a = u2/u1 shouldn't u3/u2 give the same 'r'?
  8. Aug 22, 2011 #7


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    Note, you have the formula here for an arithmetic sequence and geometric series, not sequence. This is confusing you.
  9. Aug 23, 2011 #8
    Ok, back again...

    So, we've already seen that log b - log a = d
    now, log c - log b = (a + (n + 1)d) - (a + nd)
    c - b = (a + nd + d) - (a + nd)
    c - b = d
    Thus, we see that the 3 terms of log a, b and c are all in arithmetic sequence.

    In geometric 'sequence' we have...
    Un = ar^n - 1
    so, log a = ar^n - 1, log b = ar^n and log c = ar^n + 1
    if r = U2/ U1, then log a/ log b = r
    log b/ log a = [itex]\frac{ar^n}{ar^n - 1}[/itex]
    b/ a = 1/ -1 = -1

    log c/ log b = [itex]\frac{ar^n + 1}{ar^n}[/itex]
    c/ b = 1

    Am I in the clear? Thanks, in advance.
  10. Aug 23, 2011 #9


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    You have as given by the problem statement that:

    log b - log a = log c - log b = d (for some d).

    What to you get if you take exp() of these expressions?

    This is really meant to be trivial.
  11. Aug 23, 2011 #10
    log b = d + log a, log c = d + log b = 2d + log a

    Therefore, b^2 = ac...?
    Last edited: Aug 23, 2011
  12. Aug 23, 2011 #11


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    Yes, just work it out. Just do the exponentiation. I haven't seen you try it yet.
  13. Aug 24, 2011 #12
    Ok, so if x = log a, then e^x = a.

    Thus, the exponent of log a, log b and log c is a, b and c, so...
    b = d + a and c = d + b = 2d + a
    Therefore, b^2 = ac...?
  14. Aug 24, 2011 #13


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    Do you know the basic laws of exponentiation/logs?

    exp(m + j ) = exp(m) * exp (j)

    exp(log(m)) = m

    exp( m * j) = exp(m)^j = exp(j)^m

    (deliberately using funny letters different from the OP).

    for example. Let me see you apply exp() to the expressions on both sides of the equalities you wrote in post #10.

    What is frustrating here is that actually writing out the full answer for you would be about 1/4 the length of this post. But I want you to do that.
  15. Aug 24, 2011 #14
    Unfortunately, maths was always my weakest subject and it's been a lifetime since I last studied it. My best years are behind me and I'm trying to help my grandson with his homework, but it's obviously going to be a lot harder than I imagined.

    I genuinely appreciate your assistance and wanted to say as much, as I have always found it hard to express my difficulty in the subject in a classroom environment. I always had fears of attracting negative reactions from the other students or teachers when I struggled with these concepts.

    I'm not looking for pity, but I just wanted to explain my position, as I feel maybe you think I'm frustrating you intentionally. For what it's worth, I apologise.
  16. Aug 24, 2011 #15


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    I apologize to the mentors for this, but in this case I am inclined to write out an answer:

    You had written in post #10:

    log b = d + log a, log c = d + log b = 2d + log a

    Define that d = log r ; that is r = e^d.

    Applying exp() to your equations (using the rules I gave in my previous post) then gives:

    b= a r ; c = b r; c = a r^2

    From which geometric series is obvious; also obvious that b^2 = a^2 r^2 = (a)*(a r^2) = ac.
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