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#1 2018-06-10 04:18:56

Cliff
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Sequences

(a) Determine all nonnegative integers r such that it is possible for an infinite arithmetic sequence to contain exactly r terms that are integers. Prove your answer.

(b) Determine all nonnegative integers r such that it is possible for an infinite geometric sequence to contain exactly r terms that are integers. Prove your answer.

Thank you!

#2 2018-06-10 19:18:16

bob bundy
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Registered: 2010-06-20
Posts: 8,399

Re: Sequences

hi Cliff

Welcome to the forum.

I think the answer to (a) is r=1

Here's my logic:

Certainly 1 is a possibility.

eg.  a, a + π, a + 2π + a + 3π, …. where  a is an integer and π = pi, is such a sequence.  'a' doesn't have to be the first term; you could have many terms leading up  to 'a' and then as shown.

So what about r > 1 ?

Let's say a and b are two integer terms in a sequence, m terms apart

so b - a = md where d id the common difference.  As a and b are integers, so is md.

so another integer term is c = b + md.

So if there are r such integer terms then it is always possible to find one more by adding md to the last.  So if r > 1 there will be an infinite number of such terms.

I'll post again if I find an answer to (b)

LATER EDIT:

I think r = 1 or r = 2 are the only possibilities, but I'm still working on a proof.

Bob

Last edited by bob bundy (2018-06-10 21:45:48)


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#3 2018-06-16 23:11:58

Alg Num Theory
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Registered: 2017-11-24
Posts: 338
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Re: Sequences

Bob, I think r=0 is also possible for (a).

For (b), I think the answer is all nonnegative integers. Clearly there are infinite geometric sequences with no integers, e.g.

If r = 1, add 1 as the first term to the above sequence.

If r > 1, then

is an infinite geometric sequence (common ratio 1/r) with exactly r integer terms (the first r of them).

Last edited by Alg Num Theory (2018-06-17 08:17:56)

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#4 2018-07-03 10:36:40

!nval!d_us3rnam3
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Registered: 2017-03-18
Posts: 25

Re: Sequences

Can someone work me through part (a) of this? It doesn't look like an answer has solidified.


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