Question

If $2(y-a)$ is the HM between $y-x$ and $y-z$, then $x-a,y-a,z-a$ are in.

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

Answer: (B)

GP

It is given that $y-x,2(y-a)$ and $(y-z)$ are in HP.
$\Rightarrow \frac{1}{y-x},\frac{1}{2(y-a)},\frac{1}{y-z}$ are in AP.
$\Rightarrow \frac{1}{2(y-a)}-\frac{1}{y-x}=\frac{1}{y-z}-\frac{1}{2(y-a)}$
$\Rightarrow \frac{2a-y-x}{y-x}=\frac{y+z-2a}{y-z}$
$\Rightarrow \frac{(x-a)+(y-a)}{(x-a)-(y-a)}=\frac{(y-a)+(z-a)}{(y-a)-(z-a)}$
$\Rightarrow \frac{x-a}{y-a}=\frac{y-a}{z-a}$

$\Rightarrow (x-a)(z-a)=(y-a{)}^2$
Hence, $x-a,y-a$ and $z-a$ are in GP.