Sequence Both Arithmetic and Geometric?

Algebra Level 4

Let $a_n$ denote the $n^\text{th}$ term of an infinite sequence in which the terms are not all the the same. Is it possible that $\{a_n\}$ can be both an arithmetic progression and a geometric progression?

Yes, it is possible No, never It depends on the condition

3 solutions

Michael Mendrin
Feb 2, 2017

Suppose the integer sequence ${a}_{n}$ is arithmetic. Then the sum of its reciprocals will be divergent. But if the same is geometric, then the sum of its reciprocals will be convergent! Can't have that.

I'll get back to this later with sequences other than integer. See comments below.

Many geometric sequences have a divergent sum of reciprocals. For instance, $a_n = (\tfrac12)^n$ .

Arjen Vreugdenhil - 4 years, 4 months ago

Doesn't that one converge?

Michael Mendrin - 4 years, 4 months ago

Oh, wait, maybe there's uncertainty about whether or not the sequences are integers. I assumed they were.

Michael Mendrin - 4 years, 4 months ago

It doesn't say...

Arjen Vreugdenhil - 4 years, 4 months ago

@Arjen Vreugdenhil – Right, it doesn't say. Well, my observation only applies if the sequences are strictly integers. Just an observation.

Michael Mendrin - 4 years, 4 months ago

@Michael Mendrin – It is easily saved. If the geometric common ratio is >1 (in absolute value), then consider the reciprocal series as you did. If the geometric ratio is <1 (in absolute value), then consider the sum of the sequence itself. (If the ratio is 1, then the sequence is constant and this was forbidden in the question, and if the ratio is (-1) then of course it's not arithmetic unless it's the constant 0 sequence.)

Ben Reiniger - 4 years, 4 months ago

@Ben Reiniger – What I like about this thread of conversation is that it reminds me of Cantor's Continuum Hypothesis (even though it doesn't have to do with infinite sums). But in fact, one of the unsolved problems in mathematics asks "Does every large set contain arbitrarily long arithmetic progressions?" A "large set" is an infinite sequence of integers of which the sum of their reciprocals diverges.

Michael Mendrin - 4 years, 4 months ago

Like what Ben said below, I think you should mention about the two cases when r> 1 and when r<1. Otherwise, this solution isn't complete.

Pi Han Goh - 4 years, 4 months ago

Arjen Vreugdenhil
Feb 4, 2017

Let $a,b,c$ be three successive terms of the sequence. If it is both arithmetic and geometric, $b = \frac{a+c}2 = \sqrt{ac};$ squaring and multiplying by 4 gives $a^2 + 2ac + c^2 = 4ac; \\ a^2 - 2ac + c^2 = 0; \\ (a-c)^2 = 0; \\ a = b = c.$ This means that all terms are the same, contradicting the assumption.

This is a very approach! I'm wondering if we can still solve this problem if we compare other 2 progressions (like arithmetic-geometric progression).

Pi Han Goh - 4 years, 4 months ago

Brian Charlesworth
Feb 4, 2017

Just as an algebraic alternative to Michael's clever solution ...

Suppose the first 3 terms of the sequence can be expressed both as $a, a + d, a + 2d$ and $a, ar, ar^{2}$ for some values $a,d \ne 0$ and $r \ne 1$ . Then

$(a + 2d) - (a + d) = (ar^{2} - ar) \Longrightarrow d = ar(r - 1)$ .

Now if this sequence were to extend to the next term, we would require that

$a + 3d = ar^{3} \Longrightarrow a + 3ar(r - 1) = ar^{3} \Longrightarrow a(r^{3} - 3r^{2} + 3r - 1) = 0 \Longrightarrow a(r - 1)^{3} = 0$ ,

which would only hold if either $a = 0$ or $r = 1$ , which we have ruled out. Thus if a sequence of non-constant numbers is arithmetic and geometric for 3 terms, it cannot be both for 4 terms, and thus most definitely not for an infinite number of terms.

Neat algebraic approach! It's interesting to note that we can disprove the claim by just considering the first 3 terms.

Pi Han Goh - 4 years, 4 months ago

Thanks! we can make it even stronger. If 2 terms of a sequence can be expressed as both $a, a + d$ and $a, ar$ then

$a + d = ar \Longrightarrow d = a(r - 1)$ .

To have a third term be both arithmetic and geometric we would then need

$a + 2d = ar^{2} \Longrightarrow a + 2a(r - 1) = ar^{2}$

$\Longrightarrow a(r^{2} - 2r + 1) = a(r - 1)^{2} = 0$ ,

which again would only hold if either $a = 0$ or $r = 1$ . I suppose I could make a question out of this ...

Brian Charlesworth - 4 years, 4 months ago

Sequence Both Arithmetic and Geometric?

3 solutions

0 pending reports