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Sequences and series problem 
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#1
Nov2312, 07:27 AM

PF Gold
P: 824

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_nlog_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
Nov2312, 09:14 AM

P: 295




#3
Nov2312, 12:19 PM

PF Gold
P: 824




#4
Nov2312, 02:11 PM

P: 295

Sequences and series problem
In terms of what?



#5
Nov2412, 06:58 AM

PF Gold
P: 824

a)ab b)[itex]a_n b[/itex] c)[itex]b_n  b[/itex] d)[itex]a_n  b_n [/itex] 


#6
Nov2412, 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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