### The average of n numbers x1, x2…..xn is $\overline{\boldsymbol{x}} .$ Then the value of $\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)$ s equal to.

A. n B. 0 C. n$\overline{\boldsymbol{x}}$ D. $\overline{\boldsymbol{x}}$ Answer: Option B

### Solution(By Apex Team)

According to the question, Average of ‘n’ number’s x1, x2…..xn is $\overline{\boldsymbol{x}}$ Sum of n numbers = n$\overline{\boldsymbol{x}}$ $\begin{array}{l}\therefore\large\sum_{i=1}^n\left(x_1-\overline{x}\right)\\ \text{ Put }\mathrm{i}=1,2,3\ldots..\text{ n then }\\ \left\{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\ldots..(\mathrm{xn}-\mathrm{n}\overline{x})\right\}\\ \text{ As we know that }\\ x_1+x_2+x_3+\ldots..+x_nx=n\overline{x}\\ =(n\overline{x}-n\overline{x})\\ =0\end{array}$

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