Longest geometric progression that can be obtained from a given set

[Physics lover]

Homework Statement

The longest geometric progression that can be obtained from the set
{100,101,102,....,1000}

Relevant Equations

N/A

I am searching for an easy solution to such questions.I have been playing with it for few hours.I can only make a guess because I don't know how to solve such type of questions.Although I tried assuming first term as 'a',common difference as 'r'.And then the last term that is 'ar^n-1'should be lass than 1000.And common difference should be less than 10.But it didn't helped a lot.Please help.Thanks

 

[Physics lover]

anyone?

 

[Mark44]

Homework Statement:: The longest geometric progression that can be obtained from the set
{100,101,102,...,1000}
Relevant Equations:: N/A

I am searching for an easy solution to such questions.I have been playing with it for few hours.I can only make a guess because I don't know how to solve such type of questions.

Is 100 the first term in the geometric sequence, and is 1000 the last term?
What is the significance of the numbers in between; i.e., 101, 102, ..., 999?

Although I tried assuming first term as 'a',common difference as 'r'.And then the last term that is 'ar^n-1'should be lass than 1000.And common difference should be less than 10.But it didn't helped a lot.Please help.Thanks

I think you might be confusing geometric progression and arithmetic progression. In a geometric progression, r is common ratio, not the common difference.

Example of arithmetic progression: 1, 4, 7, 10, ... Here A = 1 and d = 3
Example of geometric progression: 2, 1, 1/2, 1/4, ... Here A = 2, and r = 1/2

 

[etotheipi]

I think the idea is to create a geometric sequence with the highest possible number of terms such that each term $u_n \in \mathbb{Z} \cap [100, 1000]$. As in a possible sequence might be '$104, 260, 650$'. It seems like quite an interesting question and I'm not actually sure how to best go about it!

 

[Mark44]

I think the idea is to create a geometric sequence with the highest possible number of terms such that each term $u_n \in \mathbb{Z} \cap [100, 1000]$. As in a possible sequence might be '$104, 260, 650$'. It seems like quite an interesting question and I'm not actually sure how to best go about it!

A geometric sequence (or progression) has the form $\{ar^n \}$, where n = 0, 1, 2, ... For the sequence you show, what is the common ratio r?

 

[etotheipi]

A geometric sequence (or progression) has the form $\{ar^n \}$, where n = 0, 1, 2, ... For the sequence you show, what is the common ratio r?

For my example it's a common ratio of $2.5$. Since it wasn't stated otherwise in the question I think the implication is that any rational common ratio is permissible!

But I'm not saying my sequence is anywhere near an answer, just an example of a 'valid' sequence.

 

[Physics lover]

Is 100 the first term in the geometric sequence, and is 1000 the last term?
What is the significance of the numbers in between; i.e., 101, 102, ..., 999?
I think you might be confusing geometric progression and arithmetic progression. In a geometric progression, r is common ratio, not the common difference.

Example of arithmetic progression: 1, 4, 7, 10, ... Here A = 1 and d = 3
Example of geometric progression: 2, 1, 1/2, 1/4, ... Here A = 2, and r = 1/2

Yes,100 is the first term and 1000 is the last.Sorry for writing common difference,that was a typo.I will try to make the question a little more clear.We need to find a geometric progression whose all terms are included in the set {100,...,1000} and it has largest number of terms.

 

[etotheipi]

Are you sure this is the intended interpretation?

a geometric progression whose all terms are included in the set {100,...,1000} and it has largest number of terms.

does not necessarily imply

Yes,100 is the first term and 1000 is the last.

 

[archaic]

https://math.stackexchange.com/a/2554219

 

[Physics lover]

https://math.stackexchange.com/a/2554219

Thanks for the link.
By the way Is there any easy method to solve such questions.

 

[Mark44]

If 100 is the first number in the (geometric) progress, and 1000 is the last, then you have two equations:
$ar^0 = 100$ and $ar^n = 1000$
Solving the first gives a = 100. Substituting for a in the 2nd equation gives $100r^n = 1000$, or $r^n = 10$.

As @etotheipi said, there might not be a solution amongst the rationals for r with 1000 as the final number. You might need to fiddle with the first or last numbers in the progression for rational values of r.

 

[Physics lover]

If 100 is the first number in the (geometric) progress, and 1000 is the last, then you have two equations:
$ar^0 = 100$ and $ar^n = 1000$
Solving the first gives a = 100. Substituting for a in the 2nd equation gives $100r^n = 1000$, or $r^n = 10$.

As @etotheipi said, there might not be a solution amongst the rationals for r with 1000 as the final number. You might need to fiddle with the first or last numbers in the progression for rational values of r.

I think you did not understood the question.Okay,Let me explain it in detail.First of all,we do not need 100 as the first and 1000 as the last term in G.P.First term can be any term in between 100 and 1000 and similarly the last term is also between 100 and 1000.So we need to find such a progression which has maximum number of terms and all terms should be in this set.I think it will be clear now.

 

[Mark44]

I think you did not understood the question.Okay,Let me explain it in detail.First of all,we do not need 100 as the first and 1000 as the last term in G.P.First term can be any term in between 100 and 1000 and similarly the last term is also between 100 and 1000.So we need to find such a progression which has maximum number of terms and all terms should be in this set.I think it will be clear now.

You said this a few posts back:

Yes,100 is the first term and 1000 is the last.

I answered on the basis of what you wrote. If $a$ is different from 100, and $ar^{n-1}$ is different from 1000, you would still do what I described in my earlier post.

 

[Physics lover]

You said this a few posts back:
I answered on the basis of what you wrote. If $a$ is different from 100, and $ar^{n-1}$ is different from 1000, you would still do what I described in my earlier post.

ok.I think there is only a single method to solve such questions.

 

[SammyS]

ok.I think there is only a single method to solve such questions.

If you start with $a$ being any number from 100 to 125, and $r=2$, then the length of the geometric sequence is 4. For instance, with $a=100$ and $r=2$, the progression is: 100, 200, 400, 800. Length of 4.

If you use a rational, $r=\dfrac p q$, such that $1<r<2$ the geometric progression grows more slowly, and it's possible that you may have a longer progression. But this also complicates the task. For $r=\dfrac p q$, reduced to lowest terms, to have $a\cdot r^n$ be an integer, $a$ must have at least $n$ factors of $q$

Did you follow the link given by @archaic ? It gives a progression of length 6 starting at 128 with $r=\dfrac 3 2$.

 

[Physics lover]

If you start with $a$ being any number from 100 to 125, and $r=2$, then the length of the geometric sequence is 4. For instance, with $a=100$ and $r=2$, the progression is: 100, 200, 400, 800. Length of 4.

If you use a rational, $r=\dfrac p q$, such that $1<r<2$ the geometric progression grows more slowly, and it's possible that you may have a longer progression. But this also complicates the task. For $r=\dfrac p q$, reduced to lowest terms, to have $a\cdot r^n$ be an integer, $a$ must have at least $n$ factors of $q$

Did you follow the link given by @archaic ? It gives a progression of length 6 starting at 128 with $r=\dfrac 3 2$.

yes i saw it and that was a brilliant solution.But I think it will not work for larger sets of number.Foe eg- consider {10,11,12,...,10000}.It would take a lot of time if we go by that method.But I think there's no other choice.

 

[haruspex]

yes i saw it and that was a brilliant solution.But I think it will not work for larger sets of number.Foe eg- consider {10,11,12,...,10000}.It would take a lot of time if we go by that method.But I think there's no other choice.

[10,10000] is quite easy. Harder is something like [1000,10000].
We will always choose q=p-1, yes?
For [1000,10000] and a trial n up to 5 we can take a=1. E.g.for n=5, p=6, q=5 and qⁿ>1000.
Thereafter we can try values of n and p, assuming at first that a=1.
n=6, p=4
n=8, p=3
n=13, p=2
No point in trying n=7 because p is still 3, etc? Not sure..
For these I get the solutions:
p. a. n
4. 2. 6
3. 4. 5
2 1000 3
E.g. the first of those goes 2.3⁶, 2.3⁵.4, 2.3⁴.4² ..., 2.4⁶