## lavenderish 2 years ago Can a sequence be both arithmetic and geometric? Explain why.

1. SithsAndGiggles

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.

2. SithsAndGiggles

Except for d = 0 and r = 1, but then you just have the first term of both sequences, a.

3. KingGeorge

But that is such a sequence.

4. SithsAndGiggles

I was hoping to arrive at some contradiction, but I'm not sure how to get to it...

5. satellite73

the constant sequence is both but not a very interesting sequence if the sequence is not constant, then it cannot be both

6. SithsAndGiggles

Well there you go. I was thinking along the more "interesting" lines, I suppose.

7. KingGeorge

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.

8. SithsAndGiggles

But in general, does my reasoning still work? Given some sequences a_n and A_n.

9. KingGeorge

In general, you would have to show that if $$\large a+sd=ar^s$$ and $$a\neq 0$$, then $$d=0$$ and $$r=1$$.

10. KingGeorge

Correction: If $$\large a+sd=ar^s$$ for all $$s\in\mathbb{Z}_{\ge0}$$, $$a\neq0$$, then $$d=0$$ and $$r=1$$.

11. SithsAndGiggles

Ah, I see now. Thanks!

12. KingGeorge

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 $$0^0$$).

13. KingGeorge

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 $$sd=0\implies d=0$$.