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SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1For 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.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Except for d = 0 and r = 1, but then you just have the first term of both sequences, a.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1But that is such a sequence.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1I was hoping to arrive at some contradiction, but I'm not sure how to get to it...

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1the constant sequence is both but not a very interesting sequence if the sequence is not constant, then it cannot be both

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Well there you go. I was thinking along the more "interesting" lines, I suppose.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1There 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.

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1But in general, does my reasoning still work? Given some sequences a_n and A_n.

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1In general, you would have to show that if \(\large a+sd=ar^s\) and \(a\neq 0\), then \(d=0\) and \(r=1\).

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Correction: If \(\large a+sd=ar^s\) for all \(s\in\mathbb{Z}_{\ge0}\), \(a\neq0\), then \(d=0\) and \(r=1\).

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.1Ah, I see now. Thanks!

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1Also, 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 \(0^0\)).

KingGeorge
 one year ago
Best ResponseYou've already chosen the best response.1If \(a+sd=ar^s\), then for nonzero \(a\), \[\large \begin{aligned} \frac{sd}{a}&=r^s1\\ &=(r1)(r^{s1}+r^{s2}+...+r+1) \end{aligned} \]However, both sides must be equal for all values of \(s\ge0\), it must be that both sides are 0. Hence, \(r=1\), and \(sd=0\implies d=0\).
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