Find the variance of the sequence a, a + d, a | vedclass

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(B) The given sequence is $x_i = a + id$ for $i = 0, 1, 2, \dots, 2n$. The total number of terms is $N = 2n + 1$.The mean $\bar{x}$ is given by $\bar{

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$\frac{n(n + 1)}{3} d^2$

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(B) The given sequence is $x_i = a + id$ for $i = 0, 1, 2, \dots, 2n$. The total number of terms is $N = 2n + 1$.The mean $\bar{x}$ is given by $\bar{x} = \frac{1}{2n+1} \sum_{i=0}^{2n} (a + id) = a + d \cdot \frac{1}{2n+1} \cdot \frac{2n(2n+1)}{2} = a + nd$.The variance $\sigma^2$ is given by $\sigma^2 = \frac{1}{2n+1} \sum_{i=0}^{2n} (x_i - \bar{x})^2$.Substituting $x_i - \bar{x} = (a + id) - (a + nd) = (i - n)d$,we get:$\sigma^2 = \frac{1}{2n+1} \sum_{i=0}^{2n} (i - n)^2 d^2 = \frac{d^2}{2n+1} \sum_{j=-n}^{n} j^2$,where $j = i - n$.Since $\sum_{j=-n}^{n} j^2 = 2 \sum_{j=1}^{n} j^2 = 2 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(2n+1)}{3}$.Therefore,$\sigma^2 = \frac{d^2}{2n+1} \cdot \frac{n(n+1)(2n+1)}{3} = \frac{n(n+1)}{3} d^2$.

Find the variance of the sequence $a, a + d, a + 2d, \dots, a + 2nd$.

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