The raw data \(x_1, x_2, \ldots \ldots, x_{\mathrm{n}}\) is an A.P. with common difference \(\mathrm{d}\) and first term \(0 . \bar{x}\) and \(\sigma^2\) are mean and variance of \(x_{\mathrm{i}}, \mathrm{i}=1,2, \ldots \ldots \mathrm{n}\), then \(\sigma^2\) is

\(\begin{aligned} \bar{x}= & \frac{x_1+x_2+\ldots+x_{\mathrm{n}}}{\mathrm{n}} \\ & =\frac{\frac{\mathrm{n}}{2}\left[2 x_1+(\mathrm{n}-1) \mathrm{d}\right]}{\mathrm{n}} \\ & =\frac{(\mathrm{n}-1) \mathrm{d}}{2} \\ \sum x_{\mathrm{i}}^2 & =x_1^2+x_2^2+\ldots+x_{\mathrm{n}}^2 \\ & =0+\mathrm{d}^2+(2 \mathrm{~d})^2+\ldots+[(\mathrm{n}-1) \mathrm{d}]^2 \\ & =\mathrm{d}^2\left[1+2^2+3^2+\ldots+(\mathrm{n}-1)^2\right] \\ & =\mathrm{d}^2\left[\frac{\mathrm{n}(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}\right] \\ \sigma^2 & =\frac{1}{\mathrm{n}} \sum x_{\mathrm{i}}^2-(\bar{x})^2 \\ & =\frac{\mathrm{d}^2(\mathrm{n}-1)(2 \mathrm{n}-1)}{6}-\left[\frac{(\mathrm{n}-1) \mathrm{d}}{2}\right]^2 \\ & =\frac{\mathrm{d}^2(\mathrm{n}-1)}{2}\left(\frac{2 \mathrm{n}-1}{3}-\frac{\mathrm{n}-1}{2}\right) \\ & =\frac{\mathrm{d}^2(\mathrm{n}-1)}{2}\left(\frac{\mathrm{n}+1}{6}\right)=\frac{\left(\mathrm{n}^2-1\right) \mathrm{d}^2}{12}\end{aligned}\)