Sequence Problem

[yabbadabbadoo]

Given the sequence log3, log6, log12, ...... determine wheter the sequence is geometric or arithmetic or neither.

 

[MarkFL]

Hello, and welcome to FMH!

I would observe that we appear to have:

[MATH]a_n=\log(2^{n-1}\cdot3)[/MATH]
Can you rewrite that using logarithmic properties, and show that:

[MATH]a_n=a_1+(n-1)d[/MATH] ?

 

[Dr.Peterson]

Given the sequence log3, log6, log12, ...... determine whether the sequence is geometric or arithmetic or neither.

Another way to start would be to find the successive differences for the three given terms (that is, [MATH]a_2 - a_1[/MATH] and [MATH]a_3 - a_2[/MATH]), and see if they are equal. Then do the same with the successive ratios (that is, [MATH]a_2 / a_1[/MATH] and [MATH]a_3 / a_2[/MATH]). Then use what you see to answer the question.

 

[pka]

Given the sequence log3, log6, log12, ...... determine wheter the sequence is geometric or arithmetic or neither.

The question for me is: what is the next term in that sequence?
I see that seems natural that is should be \(\log(24)~?\)
So we have:
\(a_0=0\cdot\log(2)+\log(3)=\log(3)\\ a_1=1\cdot\log(2)+\log(3)=\log(6)\\a_2=2\cdot\log(2)+\log(3)=\log(12)\\ \vdots\\\text{In general, }a_n=n\cdot\log(2)+\log(3)=n\cdot\log\left(3\cdot 2^n\right)\)

 

[pka]

The question for me is: what is the next term in that sequence?
I see that seems natural that is should be \(\log(24)~?\)
So we have:
\(a_0=0\cdot\log(2)+\log(3)=\log(3)\\ a_1=1\cdot\log(2)+\log(3)=\log(6)\\a_2=2\cdot\log(2)+\log(3)=\log(12)\\ \vdots\\\text{In general, }a_n=n\cdot\log(2)+\log(3)=n\cdot\log\left(3\cdot 2^n\right)\)

Typo correction: \(a_n=n\cdot\log(2)+\log(3)=\log\left(3\cdot 2^n\right)\)