$\text{ The mean and variance of ' }n\text{ 'observations }{x}_1,x_2,x_3,\dots ,x_n\text{ are }6\text{ and }0,\text{ respectively. If }$$\sum _{i=1}^n x_i^2=900,\text{ then the value of }n\text{ is equal to }$

$\begin{array}{l}\text{ Given mean }\overline{x}=6\text{ and }\\ \text{ variance }{\sigma }^2=\frac{\sum _{i=1}^n {\left(x_i-\overline{x}\right)}^2}{n}=0\\ \Rightarrow {\left(x_1-6\right)}^2+{\left(x_2-6\right)}^2+\dots \dots +{\left(x_n-6\right)}^2=0\\ \Rightarrow x_1=x_2=\dots \dots ..=x_n=6\end{array}$

$\text{ ( }\because \text{ Sum of perfect squares will be zero, when each of them is zero) }$

$\begin{array}{r}\text{ Given that }\sum _{i=1}^n x_i^2=900\\ \Rightarrow n\cdot 6^2=900\\ \therefore n=25\end{array}$