Question

If $2(y-a)$ is the H.M. between $y-x$ and $y-z$, then show that $x-a,y-a,z-a$ are in G.P.

Answer 1

$(y-x),2(y-a),(y-z)$ are in H.P.
$\therefore \frac{1}{y-x},\frac{1}{2(y-a)},\frac{1}{y-z}$ are in A.P.
or $\therefore \frac{1}{2(y-a)}-\frac{1}{y-x}=\frac{1}{y-z}-\frac{1}{2(y-a)}$
or $\frac{2a-y-x}{y-x}=\frac{y+z-2a}{y-z}$
or $\frac{(x-a)+(y-a)}{(x-a)-(y-a)}=\frac{(y-a)+(z-a)}{(y-a)-(z-a)}$
Apply componendo and dividendo.
$\frac{2(x-a)}{2(y-a)}=\frac{2(y-a)}{2(z-a)}$
or $(y-a)$${}^2$ $,(y-a),(z-a)$
$\therefore$ $(x-a),(y-a),(z-a)$ are in G.P.