# Ifis the mean of the ten natural numbers

Question:

is the mean of the ten natural numbers $x_{1}, x_{2}, x_{3}, \cdots, x_{10}$ show that

$\left(\mathrm{x}_{1}-\overline{\mathrm{X}}\right)+\left(\mathrm{x}_{2}-\overline{\mathrm{X}}\right)+\cdots+\left(\mathrm{x}_{10}-\overline{\mathrm{X}}\right)=0$

Solution:

We have,

$\overline{\mathrm{x}}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}+\cdots+\mathrm{x}_{10}}{10}$

$\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}+\cdots+\mathrm{x}_{10}=10 \overline{\mathrm{x}} \cdots \cdot(1)$

Now, $\left(\mathrm{x}_{1}-\overline{\mathrm{X}}\right)+\left(\mathrm{x}_{2}-\overline{\mathrm{X}}\right)+\cdots+\left(\mathrm{x}_{10}-\overline{\mathrm{X}}\right)$

$=\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\cdots+\mathrm{x}_{10}\right)-(\overline{\mathrm{x}}+\overline{\mathrm{x}}+\overline{\mathrm{x}}+$ upto 10 terms $)$

$=\left(\mathrm{x}_{1}+\mathrm{x}_{2}+\cdots+\mathrm{x}_{10}\right)-(\overline{\mathrm{x}}+\overline{\mathrm{x}}+\overline{\mathrm{x}}+$ upto 10 terms $)$

$=10 \overline{\mathrm{x}}-10 \overline{\mathrm{x}} \quad$ [By equation(i)]

$\therefore\left(\mathrm{x}_{1}-\overline{\mathrm{X}}\right)+\left(\mathrm{x}_{2}-\overline{\mathrm{X}}\right)+\cdots+\left(\mathrm{x}_{10}-\overline{\mathrm{X}}\right)=0 .$