Sequences and series problem


by utkarshakash
Tags: sequences, series
utkarshakash
utkarshakash is offline
#1
Nov23-12, 07:27 AM
P: 655
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]
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Millennial
Millennial is offline
#2
Nov23-12, 09:14 AM
P: 295
Quote Quote by utkarshakash View Post
Then there exists [itex]x \in R^+[/itex], such that [itex]log_x a_n-log_x a[/itex] is equal to
Such that [itex]log_x a_n-log_x a[/itex] is equal to what?
utkarshakash
utkarshakash is offline
#3
Nov23-12, 12:19 PM
P: 655
Quote Quote by Millennial View Post
Such that [itex]log_x a_n-log_x a[/itex] is equal to what?
That's what I have to find.

Millennial
Millennial is offline
#4
Nov23-12, 02:11 PM
P: 295

Sequences and series problem


In terms of what?
utkarshakash
utkarshakash is offline
#5
Nov24-12, 06:58 AM
P: 655
Quote Quote by Millennial View Post
In terms of what?
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]
Millennial
Millennial is offline
#6
Nov24-12, 04:36 PM
P: 295
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.


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