1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sequences and series problem

  1. Nov 23, 2012 #1

    utkarshakash

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    Read this passage and then answer the questions that follow

    We know that, if [itex]a_1,a_2,.........,a_n[/itex] are in Harmonic Progression, then [itex]\frac{1}{a_1},\frac{1}{a_2}.........,\frac{1}{a_n},[/itex] are in Arithmetic Progression and vice versa. If [itex]a_1,a_2,.........,a_n[/itex] are in Arithmetic Progression with common difference d, then for any b(>0), the numbers [itex]b^{a_1},b^{a_2},b^{a_3},.......,b^{a_n}[/itex] are in Geometric Progression with common ratio r, then for any base b(b>0), [itex]log_b a_1,log_b a_2,...........,log_b a_n[/itex] are in Arithmetic Progression with common difference [itex]log_b r[/itex]

    Q.1. Given a Geometric Progression and an Arithmetic Progression with positive terms [itex]a,a_1,a_2,.........,a_n[/itex] and [itex]b, b_1, b_2,.............,b_n[/itex]. The common ratio of the Geometric Progression is different from 1. Then there exists [itex]x \in R^+[/itex], such that [itex]log_x a_n-log_x a[/itex] is equal to

    2. Relevant equations

    3. The attempt at a solution
    Let the common ratio of the given Geometric Progression be r.

    [itex]r= \left( \frac{a_n}{a} \right) ^{1/n}[/itex]

    Now from the last statement of the passage I can deduce that

    For [itex]x \in R^+ \\
    log_x a, log_x a_1,.......,log_x a_n[/itex]
    is in Arithmetic Progression with common difference (D) = [itex]log_x \left( \frac{a_n}{a} \right)^{1/n}[/itex]

    Let the common difference of the given Arithmetic Progression(not the above one) be d.

    [itex]d= \dfrac{b_n - b}{n}[/itex]

    Now from the second statement of the passage I can deduce that

    For [itex]x \in R^+ \\
    x^b, x^{b_1},............,x^{b_n}[/itex]
    is in Geometric Progression with common ratio (R) = [itex] x^{\frac{b_n - b}{n}}[/itex]

    I have to find [itex]log_x \dfrac{a_n}{a} \\

    nlogD=log_x \dfrac{a_n}{a}\\

    n=\dfrac{logx}{logR} (b_n - b)[/itex]

    Substituting the value of n from above into nlogD I get

    [itex]\dfrac{logx}{logR} (b_n - b) logD[/itex]
     
  2. jcsd
  3. Nov 23, 2012 #2
    Such that [itex]log_x a_n-log_x a[/itex] is equal to what?
     
  4. Nov 23, 2012 #3

    utkarshakash

    User Avatar
    Gold Member

    That's what I have to find.
     
  5. Nov 23, 2012 #4
    In terms of what?
     
  6. Nov 24, 2012 #5

    utkarshakash

    User Avatar
    Gold Member

    OK I am giving you the options

    a)a-b
    b)[itex]a_n -b[/itex]
    c)[itex]b_n - b[/itex]
    d)[itex]a_n - b_n [/itex]
     
  7. Nov 24, 2012 #6
    Depending on your choice of x, you can make it equal to a lot of things. To be more precise:

    [tex]\log_x(a_n)-\log_x(a)=\frac{\log(a_n)-\log(a)}{\log(x)}=\frac{\log(a_n/a)}{\log(x)}[/tex]

    Also, the restriction [itex]x>0[/itex] has no implications because [itex]\log(0)[/itex] is already undefined.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sequences and series problem
Loading...