The average of n numbers x1, x2.....xn is x .Then the value of _i\, = \,1^n ( x_i - x ) is equal to -

Answer: Option B

According to the question,
Average of 'n' number's x₁, x₂.....x_n is $$\overline{x}$$
Sum of n numbers = n$$\overline{x}$$
∴ $$\sum _{i=1}^n\left(x_1-\overline{x}\right)$$
Put i = 1, 2, 3.....n then
{x₁ + x₂ + x₃ + .....(xn - n$$\overline{x}$$)}
As we know that
x₁ + x₂ + x₃+.....+ x_nx = n$$\overline{x}$$
= (n$$\overline{x}$$ - n$$\overline{x}$$)
= 0