If a∈R and a1,a2,a3⋯⋯,an∈R then (x−a1)2+(x−a2)2+….+(x−an)2 assumes its least value at x =

A
$a_{1} + a_{2} + \dots \dots + a_{n}$
B
$2 (a_{1} + a_{2} + a_{3} + \dots + a_{n})$
C
$n (a_{1} + a_{2} + \dots . + a_{n})$
D
none of these

Solution:

We have,

$\begin{aligned} {(x - a_{1})}^{2} + {(x - a_{2})}^{2} + \dots \dots + {(x - a_{n})}^{2} \\ = {nx}^{2} - 2 x (a_{1} + a_{2} + \dots + a_{n}) + (a_{1}^{2} + a_{2}^{2} + \dots + a_{n}^{2}) \end{aligned}$

Clearly, $y = {nx}^{2} - 2 x (a_{1} + a_{2} + \dots . + a_{n}) + (a_{1}^{2} + a_{2}^{2} + \dots + a_{n}^{2})$

represents a parabola which opens upward. So, it attains its minimum value at the vertex ie. at

$x = \frac{2 (a_{1} + a_{2} + \dots . + a_{n})}{2 n} = \frac{a_{1} + a_{2} + \dots . + a_{n}}{n} [Using x = \frac{- b}{2 a}]$

If $a∈R$ and $a_{1}, a_{2}, a_{3 \dots \dots}, a_{n} \in R$ then ${(x - a_{1})}^{2} + {(x - a_{2})}^{2} + \dots . + {(x - a_{n})}^{2}$ assumes its least value at x =

Solution:

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