Determine the mean and standard deviation of the first $n$ terms of an $A.P.$ whose first term is $a$ and common difference is $d$.

The terms of the $A.P.$ are $x_i = a + (i-1)d$ for $i = 1, 2, \dots, n$.
$\text{Mean } (\bar{x}) = \frac{1}{n} \sum_{i=1}^{n} [a + (i-1)d] = \frac{1}{n} [na + d \frac{(n-1)n}{2}] = a + \frac{(n-1)d}{2}$.
To find the standard deviation $(\sigma)$,we use the formula $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$.
Alternatively,$\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2$.
Let $y_i = x_i - a = (i-1)d$. Then $\bar{y} = \frac{1}{n} \sum (i-1)d = \frac{d(n-1)}{2}$.
$\sum y_i^2 = d^2 \sum_{j=0}^{n-1} j^2 = d^2 \frac{(n-1)n(2n-1)}{6}$.
$\sigma^2 = \frac{1}{n} \sum y_i^2 - (\bar{y})^2 = \frac{d^2(n-1)(2n-1)}{6} - \frac{d^2(n-1)^2}{4}$.
$\sigma^2 = d^2(n-1) [\frac{2(2n-1) - 3(n-1)}{12}] = d^2(n-1) [\frac{4n-2-3n+3}{12}] = \frac{d^2(n-1)(n+1)}{12} = \frac{d^2(n^2-1)}{12}$.
Therefore,$\sigma = |d| \sqrt{\frac{n^2-1}{12}}$.