Question

If $x_i>0$, $i=1,2,3,\dots 50$and $(x_1+x_2+x_3+\dots +x_{50})=50$, then the minimum value of $\left[\frac{1}{x_1}+\frac{1}{x_2}+\dots .+\frac{1}{x_{50}}\right]$ equals

• A. $50$
• B. ${50}^2$
• C. ${50}^3$
• D. none of these

Answer 1

The correct option is A

$50$

Explanation for correct option:

Find the value of the given expression

Given sum of the series $=(x_1+x_2+x_3+\dots +x_{50})=50$

We know that, AM $\ge$ HM

or, $\frac{(a_1+a_2+a_3+\dots +a_n)}{n}\ge \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\dots .+{\displaystyle \frac{1}{a_n}}}$

where, $a_1,a_2.......a_n$are terms of any series.

So,

$\frac{(x_1+x_2+x_3+\dots +x_{50})}{50}\ge \frac{50}{\frac{1}{x_1}+\frac{1}{x_2}+\dots .+{\displaystyle \frac{1}{x_{50}}}}$

$\Rightarrow \frac{50}{50}\ge \frac{50}{\left[\frac{1}{x_1}+\frac{1}{x_2}+\dots .+\frac{1}{x_{50}}\right]}$

$\Rightarrow \left[\frac{1}{x_1}+\frac{1}{x_2}+\dots .+\frac{1}{x_{50}}\right]\ge 50$

The minimum value of $\left[\frac{1}{x_1}+\frac{1}{x_2}+\dots .+\frac{1}{x_{50}}\right]$ is $50$.

Hence, option(A), $50$ is the answer.