finding the nth term issue

jessica_r

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Mar 15, 2019
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I have 2 questions I am really struggling with related to finding the nth term....
Q1.
Sequence 1: 140,133,126...
Sequence 2: -59, -54, -49...

a) Find the nth term for each sequence and prove that the 17th term is the same on both sequences
b) there are 6 terms that are common to both sequences - can you find them all without writing out both sequences?

a)
S1: -7n+147
S2: 5n-64

therefore the 17th term is:
S1: -7(17)+147 = -119+147 = 28
S2: 5(17)-64 = 85-64 = 21

But the 17th term is not the same!

Also, i cannot find 6 terms that are common; for them to be common I've put -7n+147 = 5n-64 -> 211=12n - > n=211/12 - > but thats not an integer so not sure how to use this to find common terms - where am i going wrong?
 

Mr. Bland

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Dec 27, 2019
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Both sequences change by some constant each term, meaning their functions are both linear. The resulting slopes are not the same, so they intersect at exactly one point. The value for \(\displaystyle n\) where this intersection occurs is not a whole number, meaning it is not a term in either sequence.

Your math is sound. You have proven that a) the 17th term is not the same in both sequences and b) there are no terms common to both sequences.
 

Dr.Peterson

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Nov 12, 2017
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Technically, three terms are not enough to know the formula for a sequence, unless you are told, in this instance, that they are both arithmetic progressions. Were you? It's conceivable that they are not; but in that case you can't answer the question at all.

We can see that the 17th terms can't be the same, because the different between the sequences starts out at 199, and decreases by 12 each time. If they were ever equal, then 199 would have to be a multiple of 12, which it isn't! I'd say there's a typo or other error in the problem.

Now, part (b) could still be true. Two sequences can have terms in common for different indices -- for example, the sequences 1, 3, 5, 7, 9, 11, 13, ... and 1, 4, 7, 10, 13, ... have in common the terms 1, 7, 13, ... . Don't misread "common terms" to mean "equal at given indices"! But I'm not making any attempt to find such terms, since I don't trust the problem in the first place.

On the other hand, two arithmetic progressions can't have a finite number of terms in common! (Well, maybe in some special cases, and then it's probably always one term.) They typically behave like my example. So I doubt that (b) is valid even for whatever sequences were intended.
 

Cubist

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Oct 29, 2019
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593
A 6th order polynomial could be constructed for one of the sequences, while leaving the other one linear (for simplicity). This could be made such that 6 terms, with the same indices, match.

@jessica_r would such a solution fit in with the work that you are currently doing in class?
 

jessica_r

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Mar 15, 2019
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Th
Both sequences change by some constant each term, meaning their functions are both linear. The resulting slopes are not the same, so they intersect at exactly one point. The value for \(\displaystyle n\) where this intersection occurs is not a whole number, meaning it is not a term in either sequence.

Your math is sound. You have proven that a) the 17th term is not the same in both sequences and b) there are no terms common to both sequences.
Thank you!
 

jessica_r

New member
Joined
Mar 15, 2019
Messages
8
Technically, three terms are not enough to know the formula for a sequence, unless you are told, in this instance, that they are both arithmetic progressions. Were you? It's conceivable that they are not; but in that case you can't answer the question at all.

We can see that the 17th terms can't be the same, because the different between the sequences starts out at 199, and decreases by 12 each time. If they were ever equal, then 199 would have to be a multiple of 12, which it isn't! I'd say there's a typo or other error in the problem.

Now, part (b) could still be true. Two sequences can have terms in common for different indices -- for example, the sequences 1, 3, 5, 7, 9, 11, 13, ... and 1, 4, 7, 10, 13, ... have in common the terms 1, 7, 13, ... . Don't misread "common terms" to mean "equal at given indices"! But I'm not making any attempt to find such terms, since I don't trust the problem in the first place.

On the other hand, two arithmetic progressions can't have a finite number of terms in common! (Well, maybe in some special cases, and then it's probably always one term.) They typically behave like my example. So I doubt that (b) is valid even for whatever sequences were intended.
They are arithmetic progressions. thank you for the explanation on the the difference not being a multiple of the different values.

I did not read the question as common terms then, but rather as equal at given indices. If it is the former (so common terms where S1n =/=S2n; but how would you work that out mathamatically? I can only find 2 examples but had to work out the sequences to do that.
 

jessica_r

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Joined
Mar 15, 2019
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A 6th order polynomial could be constructed for one of the sequences, while leaving the other one linear (for simplicity). This could be made such that 6 terms, with the same indices, match.

@jessica_r would such a solution fit in with the work that you are currently doing in class?
I do not know what 6th order polynomial means to be honest; so i think that is more advanced compared to what we are doing
 

Dr.Peterson

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Nov 12, 2017
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8,887
They are arithmetic progressions. thank you for the explanation on the the difference not being a multiple of the different values.

I did not read the question as common terms then, but rather as equal at given indices. If it is the former (so common terms where S1n =/=S2n; but how would you work that out mathamatically? I can only find 2 examples but had to work out the sequences to do that.
I have made no attempt to figure out what they expect in part (b), since I don't believe the sequences are what they meant, and a solution to the wrong problem would not be helpful. In general, this would amount to solving a Diophantine equation, which I presume is not what you are learning; for a specific (valid) problem there is likely to be an easier way. (In my example, we want indices j and k such that 2j-1 = 3k-2, so that 3k - 2j = 1; or we can just find one common term, 1, and recognize that adding any multiple of 6 will yield another.)

Maybe it will help if you tell us what you have been learning, and where the problem comes from. You should also ask your teacher, or whoever supplied the problem, to find out what it is supposed to say.
 
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