Step-by-step Solution:

1. Given Series: The series is \( a, a + d, a + 2d, \ldots, a + 2nd \). 2. Calculate the Mean (\(\overline{x}\)): The mean is the average of all terms in the series: \[ \overline{x} = \frac{a + (a + d) + (a + 2d) + \ldots + (a + 2nd)}{2n + 1} \] Simplify the numerator: \[ \overline{x} = \frac{a(2n + 1) + d(1 + 2 + 3 + \ldots + 2n)}{2n + 1} \] The sum of the first \( 2n \) natural numbers is: \[ 1 + 2 + 3 + \ldots + 2n = \frac{2n(2n + 1)}{2} = n(2n + 1) \] Substitute back: \[ \overline{x} = \frac{a(2n + 1) + d \cdot n(2n + 1)}{2n + 1} \] Factor out \( (2n + 1) \): \[ \overline{x} = \frac{(2n + 1)(a + nd)}{2n + 1} = a + nd \] 3. Calculate Deviations from the Mean: The deviations from the mean \( \overline{x} = a + nd \) are: \[ x_i - \overline{x} \] For each term in the series: \[ \begin{align*} a - (a + nd) &= -nd \\ (a + d) - (a + nd) &= (1 - n)d \\ (a + 2d) - (a + nd) &= (2 - n)d \\ &\vdots \\ (a + 2nd) - (a + nd) &= nd \\ \end{align*} \] Therefore, the deviations are: \[ -nd, (1 - n)d, (2 - n)d, \ldots, 0, \ldots, (n - 2)d, (n - 1)d, nd \] 4. Evaluate the Given Expression: The expression to evaluate is: \[ \frac{nd + (n - 1)d + (n - 2)d + \ldots + 0 + d + 2d + \ldots + (n - 1)d + (n - 2)d + nd}{2n + 1} \] Simplify the numerator: \[ 2d(1 + 2 + 3 + \ldots + (n - 1) + n) \] The sum of the first \( n \) natural numbers is: \[ 1 + 2 + 3 + \ldots + n = \frac{n(n + 1)}{2} \] Therefore, the numerator becomes: \[ 2d \cdot \frac{n(n + 1)}{2} = n(n + 1)d \] Substitute back into the expression: \[ \frac{n(n + 1)d}{2n + 1} \] Therefore, the mean of the series is \( a + nd \), the deviations from the mean are \( -nd, (1 - n)d, (2 - n)d, \ldots, 0, \ldots, (n - 2)d, (n - 1)d, nd \), and the evaluated expression is: \[ \boxed{a + nd} \] The deviations from the mean are: \[ \boxed{ -nd, (1 - n)d, (2 - n)d, \ldots, 0, \ldots, (n - 2)d, (n - 1)d, nd } \] The evaluated expression is: \[ \boxed{\frac{n(n + 1)d}{2n + 1}} \]