Question

If $\frac{1}{2(y-a)}$ is the arithmetic mean of $\frac{1}{y-x}$ and $\frac{1}{y-z}$ where $a,x,y,z\in R$, then $x-a,y-a$ and $z-a$ are in

• A. A.P.
• B. G.P.
• C. A.G.P.
• D. H.P.

Answer 1

The correct option is B G.P.

Given : $\frac{1}{2(y-a)}$ is the arithmetic mean of $\frac{1}{y-x}$ and $\frac{1}{y-z}$
So, $\frac{1}{y-a}=\frac{1}{y-x}+\frac{1}{y-z}$
$\Rightarrow (y-x)(y-z)=[2y-(x+z\left)\right](y-a)\Rightarrow y^2-(x+z)y+xz=2y^2-(x+z)y-2ay+a(x+z)\Rightarrow y^2-2ay=xz-a(x+z)\Rightarrow y^2-2ay+a^2=xz-ax-az+a^2\Rightarrow (y-a{)}^2=(x-a\left)\right(z-a)$

Hence, $x-a,y-a,z-a$ are in G.P.