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Find $3$ numbers $x, y, z$ which are consecutive terms of a geometric series, if $xy$, $yz$, $zx$ and $xyz$ are consecutive terms of an arithmetic series.

OK $y=xa$ and $z=xa^2$. Also $yz=xy+b$ $zx=xy+2b$ and $xyz=xy+3b$ So by substituting we get: $a=0$ (rejected) and $a=-\frac 1 2 $. But then how do I get $b$?

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  • $\begingroup$ @Stefan: Thanks for the edit! $\endgroup$ – Alex.vollenga Nov 8 '17 at 9:20
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If $x = a$, $y = ar$ and $z = ar^2$, $(a,r \ne 0)$, then

$xy = a^2r$, $yz = a^2r^3$, $zx = a^2 r^2$ and $xyz = a^3r^3$

Since $xy$, $yz$ and $zx$ are in AP.

$$yz = \dfrac{xy + zx}{2} \iff a^2r^3 = \dfrac{a^2r + a^2r^2}{2} \implies 2r^3 = r+ r^2 \implies r(2r^2 - r - 1) = 0$$

So $r = 1, -\dfrac12, 0$.

Also, $$zx=\dfrac{xyz + yz}{2}\iff a^2r^2 = \dfrac{a^3r^3 + a^2r^3}{2} \implies 2 = ar + r \implies a = \dfrac{2 - r}{r}$$

$a = 1 (r = 1), -5(r = -1/2)$

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  • $\begingroup$ So what are the 3 numbers in each case? I assume r can't be 1, or the 3 numbers would be equal? $\endgroup$ – Alex.vollenga Nov 8 '17 at 9:32
  • $\begingroup$ Which three ? Why $r\ne 1$ ? $\endgroup$ – A---B Nov 8 '17 at 9:49
  • $\begingroup$ The numbers x, y and z. $\endgroup$ – Alex.vollenga Nov 8 '17 at 10:09
  • $\begingroup$ @Alex.vollenga For $a = r = 1$, $x = y = z = 1$. For $a = -5, r = -1/2$, $x = -5, y = 5/2, z = -5/4$ Is this what you are asking ? $\endgroup$ – A---B Nov 8 '17 at 10:11
  • $\begingroup$ @Alex.vollenga What wrong with three numbers being equal ? $1,1,1,1$ is both in AP and GP. $\endgroup$ – A---B Nov 8 '17 at 10:14

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