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utkarshakash
Gold Member
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Homework Statement
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
Homework Equations
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]