Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
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 $b$?
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, $$zx=\dfrac{xyz + yz}{2}\iff a^2r^2 = \dfrac{a^3r^3 + a^2r^3}{2} \implies 2 = ar + r \implies a = \dfrac{2 - r}{r}$$
$a = 1 (r = 1), -5(r = -1/2)$
