If the n terms a_1, a_2, ldots, a_n are in A. | vedclass

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(C) Let the terms be $a_1, a_1+r, a_1+2r, \ldots, a_1+(n-1)r$. The mean of their squares is $\frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)^2$. The square of t

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$(C)$ $\frac{r^2(n^2-1)}{12}$

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(C) Let the terms be $a_1, a_1+r, a_1+2r, \ldots, a_1+(n-1)r$. The mean of their squares is $\frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)^2$. The square of their mean is $\left(\frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)\right)^2$. The difference is the variance of the $A$.$P$.,which is given by $\sigma^2 = \frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)^2 - \left(\frac{1}{n} \sum_{k=0}^{n-1} (a_1+kr)\right)^2$. Using the formula for the variance of the first $n$ terms of an $A$.$P$.,$\sigma^2 = \frac{r^2(n^2-1)}{12}$. Thus, the difference is $\frac{r^2(n^2-1)}{12}$.

If the $n$ terms $a_1, a_2, \ldots, a_n$ are in $A$.$P$. with common difference $r$, then the difference between the mean of their squares and the square of their mean is

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