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lavenderish

Can a sequence be both arithmetic and geometric? Explain why.

  • one year ago
  • one year ago

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  1. SithsAndGiggles
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    For an arithmetic sequence a_n, each successive term has a common difference d. For a geometric sequence b_n, each successive term has a common ratio r. For a_n, the terms look like \[\left\{a, a+d,a+2d,\ldots\right\}\] and for A_n, you have \[\left\{a,ar, ar^2,\ldots\right\}\] Suppose a_n = A_n. That means that every term of a_n corresponds exactly with every term of A_n, meaning \[a_1=A_1,\\ a_2=A_2,\\ \vdots\] Under this assumption, you have \[\begin{cases}a = a\\ a+d=ar\\ a+2d=ar^2\\ \vdots\end{cases}\] I'm pretty sure there are no real numbers d and r that satisfy the system, so I don't think a sequence can be both arithmetic and geometric.

    • one year ago
  2. SithsAndGiggles
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    Except for d = 0 and r = 1, but then you just have the first term of both sequences, a.

    • one year ago
  3. KingGeorge
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    But that is such a sequence.

    • one year ago
  4. SithsAndGiggles
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    I was hoping to arrive at some contradiction, but I'm not sure how to get to it...

    • one year ago
  5. satellite73
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    the constant sequence is both but not a very interesting sequence if the sequence is not constant, then it cannot be both

    • one year ago
  6. SithsAndGiggles
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    Well there you go. I was thinking along the more "interesting" lines, I suppose.

    • one year ago
  7. KingGeorge
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    There is no contradiction, since there are example of sequences that are both arithmetic and geometric, such as 0,0,... a sequence can be both arithmetic and geometric. However, as satellite said, this isn't a very interesting sequence.

    • one year ago
  8. SithsAndGiggles
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    But in general, does my reasoning still work? Given some sequences a_n and A_n.

    • one year ago
  9. KingGeorge
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    In general, you would have to show that if \(\large a+sd=ar^s\) and \(a\neq 0\), then \(d=0\) and \(r=1\).

    • one year ago
  10. KingGeorge
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    Correction: If \(\large a+sd=ar^s\) for all \(s\in\mathbb{Z}_{\ge0}\), \(a\neq0\), then \(d=0\) and \(r=1\).

    • one year ago
  11. SithsAndGiggles
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    Ah, I see now. Thanks!

    • one year ago
  12. KingGeorge
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    Also, I only include \(a\neq0\) since if \(a=0\), then all we need is \(d=0\) to guarantee us the sequence \(0,0,...\) (assuming you define \(0^0\)).

    • one year ago
  13. KingGeorge
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    If \(a+sd=ar^s\), then for non-zero \(a\), \[\large \begin{aligned} \frac{sd}{a}&=r^s-1\\ &=(r-1)(r^{s-1}+r^{s-2}+...+r+1) \end{aligned} \]However, both sides must be equal for all values of \(s\ge0\), it must be that both sides are 0. Hence, \(r=1\), and \(sd=0\implies d=0\).

    • one year ago
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