What is the Relationship Between Logarithms and Progressions in Mathematics?

[utkarshakash]

Homework Statement

Read this passage and then answer the questions that follow

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

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

Homework Equations

The Attempt at a Solution

Let the common ratio of the given Geometric Progression be r.

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

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

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

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

d= \dfrac{b_n - b}{n}

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

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

I have to find log_x \dfrac{a_n}{a} \\<br /> <br /> nlogD=log_x \dfrac{a_n}{a}\\<br /> <br /> n=\dfrac{logx}{logR} (b_n - b)

Substituting the value of n from above into nlogD I get

\dfrac{logx}{logR} (b_n - b) logD

 

[Millennial]

Then there exists x \in R^+, such that log_x a_n-log_x a is equal to

Such that log_x a_n-log_x a is equal to what?

 

[utkarshakash]

Such that log_x a_n-log_x a is equal to what?

That's what I have to find.

 

[Millennial]

In terms of what?

 

[utkarshakash]

In terms of what?

OK I am giving you the options

a)a-b
b)a_n -b
c)b_n - b
d)a_n - b_n

 

[Millennial]

Depending on your choice of x, you can make it equal to a lot of things. To be more precise:

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

Also, the restriction x&gt;0 has no implications because \log(0) is already undefined.