Question

If $y-z\propto \frac{1}{x},z-x\propto \frac{1}{y}$ and $x-y\propto \frac{1}{z}$, sum of three variation constants is

• A. $2$
• B. $1$
• C. $-1$
• D. $0$

Answer 1

The correct option is D $0$

Given:
$y-z\propto \frac{1}{x},z-x\propto \frac{1}{y}\text{and}x-y\propto \frac{1}{z}$
$y-z=k\frac{1}{x}.eq\left(1\right)$
$z-x=m\frac{1}{y}\dots eq\left(2\right)$
$x-y=n\frac{1}{z}..eq\left(3\right)$
Where $k,m$ and $n$ are constants,
Adding all the three equations we get,
$y-z+z-x+x-y=(\frac{k}{x}+\frac{m}{y}+\frac{n}{z})$
$\Rightarrow 0=(\frac{k}{x}+\frac{m}{y}+\frac{n}{z})$
Now, $k,m$, and $n$ are also constants.
Then, the sum of these variation constants will also be 0