1 and 20 are first and last terms of the arithmetic progression. If all the terms of this arithmetic progression are integers, then find the different number of terms that this arithmetic progression can have ?
Number of terms in an Arithmetic progression
2
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sequences-and-series
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0Sounds like homework. Is it? At any rate, here's a hint. An arithmetic sequence always has the form a, a + n, a + 2n, ... . You know that a = 1 and 20 = a + nk. Go from there. – 2012-06-13
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2Let $s$ be the step size and $k$ be the number of terms. Then $1+s(k-1)=20$. Simplify. – 2012-06-13
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0Hi, my doubt is that will you consider : 1 and 20 as a arithmetic progression. – 2012-06-13
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0@ArpitBajpai Yes, with step size 19. – 2012-06-13
1 Answers
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Here's an obvious generalization which may be interesting to you, and which exposes the reasoning for the question above: $1$ and $(1+p)$ are the first and last terms of an arithmetic progression, where $p$ is prime. If all the terms of this arithmetic progression are integers, how many different number of terms can this arithmetic progression have?
See if you can generalize Henning's comment regarding your original question which was that the number of terms $k$ with step size $s$ must satisfy $1+s(k-1)=20$ in order to answer this question as well.
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0That is 2 and p + 1. – 2012-06-13
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0but my confusion is that what should be minimum number of terms in an arithmetic progression. – 2012-06-13
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0@ArpitBajpai You are correct, 2 and p+1 is the answer. 2 terms still qualify as an arithmetic progression. – 2012-06-13
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0@Eugene The comments provided by others sufficiently addressed the OP's question, so to prevent this question from remaining unanswered I will leave this as an answer and make it community wiki, which is hopefully satisfactory. – 2012-06-13
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0@MichaelBoratko Oh. Yup you're absolutely right about that. – 2012-06-13