JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 25)
Explanation
$$\begin{aligned} &\begin{aligned} & \operatorname{Var}(8,21,34,47, \ldots \ldots, 320) \\ & \operatorname{Var}(0,13,26,39, \ldots \ldots, 312) \\ & 13^2 \cdot \operatorname{Var}(0,1,2, \ldots \ldots, 24) \\ & 13^2 \cdot \operatorname{Var}(1,2,3, \ldots \ldots, 25) \end{aligned}\\ &\text { So, } \sigma^2=13^2 \times\left(\frac{25^2-1}{12}\right)=8788\\ &\text { Alternate solution }\\ &\begin{aligned} & 8+(n-1) 13=320 \\ & 13 n=325 \\ & n=25 \end{aligned} \end{aligned}$$
$$\begin{aligned} &\text { no. of terms }=25\\ &\begin{aligned} & \text { mean }=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}=\frac{8+21+\ldots+320}{25}=\frac{\frac{25}{2}(8+320)}{25} \\ & \text { variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}}^2}{\mathrm{n}}-(\text { mean })^2 \\ & =\frac{8^2+21^2+\ldots .+320^2}{13}-(164)^2 \\ & =8788 \end{aligned} \end{aligned}$$
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