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**Cliff****Guest**

(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!

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,354

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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**Alg Num Theory****Member**- Registered: 2017-11-24
- Posts: 224
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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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