Sequences and series problem

[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 $x \in R^+ \\ log_x a, log_x a_1,...,log_x a_n$
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 $x \in R^+ \\ x^b, x^{b_1},...,x^{b_n}$
is in Geometric Progression with common ratio (R) = $x^{\frac{b_n - b}{n}}$

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

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 $x>0$ has no implications because $\log(0)$ is already undefined.