QJEEM25Mac17e: JEE Main 2025 Mathematics - Class 11 JEEM25Mac17e Solution: The variance of the numbers is _______.

$\begin{aligned}&\begin{aligned}& \operatorname{Var}(8,21,34,47, \ldots \ldots, 320) \\& \operatorname{Var}(0,13,26,39, \ldots \ldots, 312) \\& 13^2 \cdot \operatorname{Var}(0,1,2, \ldots \ldots, 24) \\& 13^2 \cdot \operatorname{Var}(1,2,3, \ldots \ldots, 25)\end{aligned}\\&\text { So, } \sigma^2=13^2 \times\left(\frac{25^2-1}{12}\right)=8788\\&\text { Alternate solution }\\&\begin{aligned}& 8+(n-1) 13=320 \\& 13 n=325 \\& n=25\end{aligned}\end{aligned}$

$\begin{aligned}&\text { no. of terms }=25\\&\begin{aligned}& \text { mean }=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{8+21+\ldots+320}{25}=\frac{\frac{25}{2}(8+320)}{25} \\& \text { variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{\mathrm{n}}-(\text { mean })^2 \\& =\frac{8^2+21^2+\ldots .+320^2}{13}-(164)^2 \\& =8788\end{aligned}\end{aligned}$