Question

If $2\left(y-a\right)$ is the harmonic mean between $y-x$ and $y-z$ , then $x-a,y-a$ and $z-a$ are in

• A. arithmetic progression
• B. geometric progression
• C. harmonic progression
• D. none of these

Answer 1

Answer: (B)

It is given that, $y-x,2\left(y-a\right)and\left(y-z\right)$ are in harmonic progression.
$\Rightarrow \frac{1}{y-x},\frac{1}{2\left(y-a\right)},\frac{1}{y-z}$ are in arithmetic progression.
$\Rightarrow \frac{1}{2\left(y-a\right)}-\frac{1}{y-x}=\frac{1}{y-z}-\frac{1}{2\left(y-a\right)}$
$\Rightarrow$ $\frac{2a-y-x}{y-x}=\frac{y+z-2a}{y-z}$
$\Rightarrow \frac{\left(x-a\right)+\left(y-a\right)}{\left(x-a\right)-\left(y-a\right)}=\frac{\left(y-a\right)+\left(z-a\right)}{\left(y-a\right)-\left(z-a\right)}$
$\Rightarrow$ $\frac{x-a}{y-a}=\frac{y-a}{z-a}$
Hence, $x-a,y-aandz-a$ are in geometric progression