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If M is the mean of x1, x2,x3,x4,x5 and x6, Prove that

Question:

If $M$ is the mean of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ and $x_{6}$, Prove that $\left(x_{1}-M\right)+\left(x_{2}-M\right)+\left(x_{3}-M\right)+\left(x_{4}-M\right)+\left(x_{5}-M\right)+\left(x_{6}-M\right)=0$.

Solution:

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

 

 

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