# Sequences and series problem

by utkarshakash
Tags: sequences, series
 P: 638 1. The problem statement, all variables and given/known data 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 2. Relevant equations 3. 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 $\dfrac{logx}{logR} (b_n - b) logD$
P: 295
 Quote by utkarshakash 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?
P: 638
 Quote by Millennial Such that $log_x a_n-log_x a$ is equal to what?
That's what I have to find.

P: 295

## Sequences and series problem

In terms of what?
P: 638
 Quote by Millennial In terms of what?
OK I am giving you the options

a)a-b
b)$a_n -b$
c)$b_n - b$
d)$a_n - b_n$
 P: 295 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.

 Related Discussions Calculus & Beyond Homework 44 Precalculus Mathematics Homework 3 Calculus & Beyond Homework 1 Calculus & Beyond Homework 1 Precalculus Mathematics Homework 4