If $$a_1,a_2,a_3,......a_n\space ∈ R,$$​, then $$(x-a_1)^2+(x-a_2)^2+.......+(x-a_n)^2=0$$​ assumes its least value at:

Correct option is A

Solution:

To find the value of  x  that minimizes the function:

(x - a₁)² + (x - a₂)² + ...... + (x - aₙ)²
The given function represents the sum of squared deviations from a set of real numbers a₁, a₂, ..., aₙ.
This type of function is minimized when x is the mean (average) of the given values.

The function
f(x) = Σ (x - aᵢ)² for i = 1 to n
is minimized when x is the mean of the given values, which is:
​$$x =\frac{ (a₁ + a₂ + ... + aₙ) }{ n}$$​​

Thus, the given function assumes its least value at:
​$$x =\frac{ (a₁ + a₂ + ... + aₙ) }{ n}$$​​

which is the arithmetic mean of a₁, a₂, ..., aₙ.