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anonymous
 3 years ago
Can a sequence be both arithmetic and geometric? Explain why.
anonymous
 3 years ago
Can a sequence be both arithmetic and geometric? Explain why.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For 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.

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I was hoping to arrive at some contradiction, but I'm not sure how to get to it...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the constant sequence is both but not a very interesting sequence if the sequence is not constant, then it cannot be both

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well there you go. I was thinking along the more "interesting" lines, I suppose.

KingGeorge
 3 years 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0But in general, does my reasoning still work? Given some sequences a_n and A_n.

KingGeorge
 3 years 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
 3 years 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 \(r=1\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, I see now. Thanks!

KingGeorge
 3 years 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
 3 years 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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