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    a) show that Logam, Logamn and Logamn2 are 3 consecutive terms of an arithmetic sequence whose common difference is Logan.

    b) Given that mn2=a, show that the sum of the first 5 terms of the arithmetic sequence with the first term Logam and common difference Logan is 5.

    I can do a) but I am not sure of b).

    Thanks (I also know that it involves the sum of the n terms formula but I cannot apply it in this unfamiliar situation)
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    EDIT: Sorry, my bad.

    What is the third term of the sequence in terms of m and n?

    How can you write each of the 5 terms in term in terms of the third term and the common difference of the sequence?
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    [QUOTE=aznkid66;42111395]EDIT: Sorry, my bad.

    What is the third term of the sequence in terms of m and n?

    How can you write each of the 5 terms in term in terms of the third term and the common difference of the sequence?[/QUOTE
    I am not actually too sure Do you mind explaining? I am so bad at sequences at series

    Thanks so much!
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    (Original post by aznkid66)
    EDIT: Sorry, my bad.

    What is the third term of the sequence in terms of m and n?

    How can you write each of the 5 terms in term in terms of the third term and the common difference of the sequence?
    What do you mean? The third term is just log mn^2 = log a

    So we are given mn^2 = a. So the first term is:

    log (a/(n^2))

    Work out the rest of the terms and then add them together

    EDIT: woops, I thought you were the OP. Thought you were asking questions rather than testing :P
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    Okay, let's start with finding the third term. You kind of already did this working in part (a), but basically you want to do:

    u_3=u_1+2d

    Then substitute the given values for u1 and d.

    As for representing each term u1, u2, u3, u4, u5 in terms of u3 and d, think: How far away from u3 is u1? How far away from u3 is u2? etc.
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    Indeed, as 2710 implied, you don't have to do the second step I suggested, but it makes calculations easier. If you can't see it, let's use a concrete yet simpler sequence:

    1, 3, 5, 7, 9

    This is the same as

    5-4, 5-2, 5, 5+2, 5+4

    And if we sum it up, we get:

    5+5+5+5+5 -2+2 -4+4

    See what I did there?

    Also, it follows from the first-last formula of an arithmetic series.

    S_n=n \times \dfrac{(u_1+u_n)}{2}

    And if n is odd:

    \dfrac{u_1+u_n}{2}=u_{(\frac{n+1  }{2})}

    Thus, again only if n is odd:

    S_n = n \times u_{(\frac{n+1}{2})}
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    (Original post by upthegunners)
    a) show that Logam, Logamn and Logamn2 are 3 consecutive terms of an arithmetic sequence whose common difference is Logan.

    b) Given that mn2=a, show that the sum of the first 5 terms of the arithmetic sequence with the first term Logam and common difference Logan is 5.

    I can do a) but I am not sure of b).

    Thanks (I also know that it involves the sum of the n terms formula but I cannot apply it in this unfamiliar situation)
    Look at it this way:

    We know the first three terms:

    log m , log mn, log mn^2 (and you should know the last two as well)

    We are given that mn^2 = a

    log m is therefore = log (a/(n^2))

    Do the same for the next 4 terms, and then add them all together
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    (Original post by aznkid66)
    Okay, let's start with finding the third term. You kind of already did this working in part (a), but basically you want to do:

    u_3=u_1+2d

    Then substitute the given values for u1 and d.

    As for representing each term u1, u2, u3, u4, u5 in terms of u3 and d, think: How far away from u3 is u1? How far away from u3 is u2? etc.
    Okay so u3 = mn^2 + 2(loga^n)
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    Surely you're making it overcomplicated?

    You've been able to show part a.

    For part b, you know the sum is:

    S_n = \dfrac {n}{2} [ 2a +(n-1)d]

    if you subbed in 5 for n, log_a m for a, and log_a n for d, what equation would you have?
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    (Original post by upthegunners)
    Okay so u3 = mn^2 + 2(loga^n)
    Um... not exactly? ^^;

    The first term is log(m), not mn^2.

    In addition, as 2710 pointed out, it might be helpful to note that in this case part (a) and part (b) feature the very same sequence!
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    (Original post by 2710)
    Look at it this way:

    We know the first three terms:

    log m , log mn, log mn^2 (and you should know the last two as well)

    We are given that mn^2 = a

    log m is therefore = log (a/(n^2))

    Do the same for the next 4 terms, and then add them all together

    log m , log mn, log mn^2
    Spoiler:
    Show
    how do we know those are the first 3 terms?
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    (Original post by upthegunners)
    log m , log mn, log mn^2
    Spoiler:
    Show
    how do we know those are the first 3 terms?
    Its continued from the first question...

    And even if u didnt have the first question, it states:

    log m is the first term, and the common difference is log n

    so log m, logm + log n = log mn , log mn + log n = log mn^2
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    (Original post by aznkid66)
    Um... not exactly? ^^;

    The first term is log(m), not mn^2.

    In addition, as 2710 pointed out, it might be helpful to note that in this case part (a) and part (b) feature the very same sequence!
    it said a=mn^2
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    This may or may not be tripping you up, but in this question the constant 'a' represents the base of the logarithm and not the first term. XD
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    (Original post by aznkid66)
    This may or may not be tripping you up, but in this question the constant 'a' represents the base of the logarithm and not the first term. XD
    Yeh I think thats wuts triiping him up. a is not the first term! lol.
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    (Original post by aznkid66)
    This may or may not be tripping you up, but in this question the constant 'a' represents the base of the logarithm and not the first term. XD
    Yeap, that's what it was haha

    I thought it was a as in a = the first term

    They could have chosen a better notation
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    Sum of an arithmetic progression:

    \displaystyle S_n=\frac{n}{2}\left(2a_1 + (n-1)d\right)

    You don't need to find the 3rd term.
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    I see no one likes my method.

    Yeah, fine, use your first-difference formula and logarithm rules. XD
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    (Original post by upthegunners)
    a) show that Logam, Logamn and Logamn2 are 3 consecutive terms of an arithmetic sequence whose common difference is Logan.

    b) Given that mn2=a, show that the sum of the first 5 terms of the arithmetic sequence with the first term Logam and common difference Logan is 5.

    I can do a) but I am not sure of b).

    Thanks (I also know that it involves the sum of the n terms formula but I cannot apply it in this unfamiliar situation)
    b) formula for sum of terms

    n=5, a= logam, d = logan

    Sn= 5/2 [2logam + (5-1)logan]

    = 5/2[2logam + 4logan]

    = 5/2[logam2 + logan4]

    = 5/2 [loga(m2n4)] (note: m2n4 =a2 since a=mn2)
    =5/2 [logaa2]
    = 5/2 [2logaa] (note: logaa = 1)
    = 5/2 x 2
    = 5
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    (Original post by Thamesgbaski)
    b) formula for sum of terms

    n=5, a= logam, d = logan

    Sn= 5/2 [2logam + (5-1)logan]

    = 5/2[2logam + 4logan]

    = 5/2[logam2 + logan4]

    = 5/2 [loga(m2n4)] (note: m2n4 =a2 since a=mn2)
    =5/2 [logaa2]
    = 5/2 [2logaa] (note: logaa = 1)
    = 5/2 x 2
    = 5
    thanks very much

    brilliant response
 
 
 
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