For (b), I think the answer is all nonnegative integers. Clearly there are infinite geometric sequences with no integers, e.g.
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If r = 1, add 1 as the first term to the above sequence.
If r > 1, then
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is an infinite geometric sequence (common ratio 1/r) with exactly r integer terms (the first r of them).
]]>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
]]>(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!
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