If $M$ is the mean of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ and $x_{6}$, Prove that $\left(x_{1}-M\right)+\left(x_{2}-M\right)+\left(x_{3}-M\right)+\left(x_{4}-M\right)+\left(x_{5}-M\right)+\left(x_{6}-M\right)=0$.
Let $M$ be the mean of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ and $x_{6}$
Then,
$\mathrm{M}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}+\mathrm{x}_{5}+\mathrm{x}_{6}}{6}$
$=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}=6 M$
To Prove :- $\left(x_{1}-M\right)+\left(x_{2}-M\right)+\left(x_{3}-M\right)+\left(x_{4}-M\right)+\left(x_{5}-M\right)+\left(x_{6}-M\right)=0$
Proof: L. H. S
$=\left(x_{1}-M\right)+\left(x_{2}-M\right)+\left(x_{3}-M\right)+\left(x_{4}-M\right)+\left(x_{5}-M\right)+\left(x_{6}-M\right)$
$=\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)-(M+M+M+M+M+M)$
= 6M – 6M
= 0
= R.H.S