Question

If $x_1,x_2,.....,x_n$ are an observation such that $\sum _{i=1}^n{x}_i^2=400$ and $\sum _{i=1}^n{x}_i=80$ ,then the least value of $n$ is

• A. 18
• B. 12
• C. 15
• D. 16

Answer 1

The correct option is D 16

Using AM $\ge$ GM, $\frac{(x_1+x_2+x_3+....+x_n)}{n}\ge (x_1\times x_2\times x_3...\times x_n{)}^{1/n}$

$\therefore \frac{80}{n}\ge (x_1\times x_2\times x_3...\times x_n{)}^{1/n}$

Squaring both sides, we have,

$\therefore \frac{6400}{n^2}\ge (x_1\times x_2\times x_3...\times x_n{)}^{2/n}-(1)$

$\frac{(x_1^2+x_2^2+x_3^3+....+x_n^2)}{n}\ge (x_1\times x_2\times x_3...\times x_n{)}^{3/n}$

$\therefore \frac{400}{n}\ge (x_1\times x_2\times x_3...\times x_n{)}^{2/n}-(2)$

For minimum value of $n$ equality exists.

So, $\frac{6400}{n^2}=\frac{400}{n}=(x_1\times x_2\times x_3...\times x_n{)}^{2/n}$

$\therefore n=16$