First slide

Theory of expressions

Question

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 =

Moderate

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

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