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SithsAndGigglesBest ResponseYou've already chosen the best response.1
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

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

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

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

satellite73Best ResponseYou've already chosen the best response.1
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

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

KingGeorgeBest ResponseYou've already chosen the best response.1
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

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

KingGeorgeBest ResponseYou've already chosen the best response.1
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

KingGeorgeBest ResponseYou've already chosen the best response.1
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

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

KingGeorgeBest ResponseYou've already chosen the best response.1
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

KingGeorgeBest ResponseYou've already chosen the best response.1
If \(a+sd=ar^s\), 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 \(sd=0\implies d=0\).
 one year ago
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