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 ?

  • $\begingroup$ Sounds 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. $\endgroup$ – Rick Decker Jun 13 '12 at 16:19
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    $\begingroup$ Let $s$ be the step size and $k$ be the number of terms. Then $1+s(k-1)=20$. Simplify. $\endgroup$ – Henning Makholm Jun 13 '12 at 16:20
  • $\begingroup$ Hi, my doubt is that will you consider : 1 and 20 as a arithmetic progression. $\endgroup$ – Arpit Bajpai Jun 13 '12 at 16:40
  • $\begingroup$ @ArpitBajpai Yes, with step size 19. $\endgroup$ – process91 Jun 13 '12 at 17:01

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.

  • $\begingroup$ That is 2 and p + 1. $\endgroup$ – Arpit Bajpai Jun 13 '12 at 17:08
  • $\begingroup$ but my confusion is that what should be minimum number of terms in an arithmetic progression. $\endgroup$ – Arpit Bajpai Jun 13 '12 at 17:09
  • $\begingroup$ @ArpitBajpai You are correct, 2 and p+1 is the answer. 2 terms still qualify as an arithmetic progression. $\endgroup$ – process91 Jun 13 '12 at 17:11
  • $\begingroup$ @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. $\endgroup$ – process91 Jun 13 '12 at 17:12
  • $\begingroup$ @MichaelBoratko Oh. Yup you're absolutely right about that. $\endgroup$ – Eugene Jun 13 '12 at 18:32

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