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If the i-th, j-th, and k-th terms in an arithmetic progression are in a geometric progression with ratio r, find r in terms of i, j, and k.

This is my result:

(1) if $ik \ne j^2$ then $r=\frac{k-j}{j-i}$; (2) if $ik = j^2$ then $r=\frac{j}{i}$.

I solved this here (Arithmetic and Geometric Progression Question 1) but I feel my proof is awkward, and my hope is that someone can come up with a more elegant solution.

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  • $\begingroup$ As a small point, in the second case, $j/i = k/j$, so you can rewrite the answer to look like the ratio of case 1...but without the subtracted bits. This didn't lead me to any further insight, however. :( $\endgroup$ Jul 6, 2015 at 2:29

2 Answers 2

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First, note that your answer's second case is redundant. If $j^2=ik$ then $$\frac{k}{j}=\frac{j}{i} = \frac{k-j}{j-i}$$ when $j\neq i$. (Note that if $i=0$ or $j=0$ then one of the others must be zero, so $i,j,k$ are not distinct in that case.)

Now, if $a,b,c$ are distinct and in geometric progression, then they must be non-zero, so $$r=\frac{b}{a}=\frac{c}{b}=\frac{c-b}{b-a}$$

If they are in arithmetic progression also, with indices $i,j,k$, then:

$$\frac{c-b}{b-a}=\frac{k-j}{j-i}$$

So the heart of this is the basic result:

$$\frac{a}{b}=\frac{c}{d},c\neq d\implies\frac{a-c}{b-d}=\frac{a}{b}.$$

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  • $\begingroup$ Nice! But it doesn't handle the case where $b = 0$, or where $j$ is the average of $i$ and $k$ (in which case I think $b = 0$ is also true). $\endgroup$ Jul 6, 2015 at 2:41
  • $\begingroup$ If $b=0$ then the progression is a constant sequence and $r=1$. $\endgroup$
    – ajotatxe
    Jul 6, 2015 at 2:45
  • $\begingroup$ If $b=0$ then $r=1$, and $i,j,k$ can be any values, which makes the case uninteresting. @JohnHughes $\endgroup$ Jul 6, 2015 at 2:56
  • $\begingroup$ You chose to divide by i+k-2j, so you need to consider what happens when this is zero. $\endgroup$ Jul 6, 2015 at 3:48
  • $\begingroup$ @martycohen true. If that is zero, then $j^2-ik\neq 0$ or $i=j=k$ by AM/GM. So then $b=0$, which we've excluded. $\endgroup$ Jul 6, 2015 at 3:52
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I didn't get the bifurcation by cases, so maybe I have an error. Here is my calculation anyway, for what it is worth:

For simplicity, let A be the $j^{th}$ term and P the arithmetic period. Then the $i^{th}$ term is A + (i-j)P and the $k^{th}$ term is A + (k-j)P. The assumtion then implies that $$ \frac{A}{A+(i-j)P} = \frac{A+(k-j)P}{A}$$

Either of these quotients give the geometric ratio you seek. Cross multiply and simplify to get $$ 0 = (k + i - 2j)A + (i-j)P \Rightarrow P = \frac{2j-k-i}{(i-j)(k-j)}A$$

To get the desired ratio substitute this expression for P into one of the (equivalent) quotients which gives the ratio. Using the right hand quotient we quickly get that the desired ratio is $$1 + \frac{2j - k-i}{i-j} = \frac{j-k}{i-k}=\frac{k-j}{j-i}$$ Not sure where the bifurcation might arise....

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  • $\begingroup$ I think the bifurcation might apply if A=0. $\endgroup$ Jul 6, 2015 at 3:52

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