Mean Calculation for Arithmetic Progression

The problem involves a sequence of numbers forming an arithmetic progression (AP): 1, 1 + d, 1 + 2d, ..., 1 + 100d. We are asked to find the value of d, given that the mean deviation from the mean is 255.

First, let's determine the mean of this arithmetic progression. The sequence consists of $n = 101$ terms. The first term is $a_1 = 1$, and the last term is $a_{101} = 1 + 100d$. The mean ($\bar{x}$) of an AP can be found by averaging the first and the last term:

$$ \bar{x} = \frac{a_1 + a_{101}}{2} $$

Substituting the given values:

$$ \bar{x} = \frac{1 + (1 + 100d)}{2} = \frac{2 + 100d}{2} = 1 + 50d $$

An alternative method confirms this: since there are 101 terms (an odd number), the mean is the middle term. The middle term is the $\frac{101+1}{2} = 51$st term. The $k$-th term of an AP is $a_k = a_1 + (k-1)d$. Thus, the 51st term is:

$$ a_{51} = 1 + (51-1)d = 1 + 50d $$

Hence, the mean of the sequence is $\bar{x} = 1 + 50d$.

Calculating Deviations from the Mean

Next, we calculate the deviation of each term ($x_i$) from the mean ($\bar{x}$). The terms are $x_i = 1 + (i-1)d$ for $i = 1, 2, ..., 101$. The deviation is $x_i - \bar{x}$:

$$ x_i - \bar{x} = (1 + (i-1)d) - (1 + 50d) $$

$$ x_i - \bar{x} = (i - 1 - 50)d $$

$$ x_i - \bar{x} = (i - 51)d $$

The deviations range from $(1-51)d = -50d$ for the first term to $(101-51)d = 50d$ for the last term. The sequence of deviations is $\{-50d, -49d, ..., -d, 0, d, ..., 49d, 50d\}$.

Sum of Absolute Deviations Calculation

The mean deviation requires the sum of the absolute values of these deviations. The absolute deviations are $|(i - 51)d|$, which form the set $\{50|d|, 49|d|, ..., |d|, 0, |d|, ..., 49|d|, 50|d|\}$.

The sum of these absolute deviations is:

$$ \sum_{i=1}^{101} |x_i - \bar{x}| = |0| + \sum_{i=1}^{50} |(i - 51)d| + \sum_{i=52}^{101} |(i - 51)d| $$

This simplifies to:

$$ = 0 + 2 \sum_{k=1}^{50} k|d| \quad (\text{due to symmetry}) $$

$$ = 2|d| \sum_{k=1}^{50} k $$

Using the formula for the sum of the first $m$ natural numbers, $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$:

$$ \sum_{k=1}^{50} k = \frac{50(50+1)}{2} = \frac{50 \times 51}{2} = 25 \times 51 = 1275 $$

Therefore, the sum of the absolute deviations is:

$$ \text{Sum} = 2|d| \times 1275 = 2550 |d| $$

Mean Deviation Calculation and Solving for d

The Mean Deviation (MD) is calculated by dividing the sum of absolute deviations by the total number of terms ($n=101$):

$$ MD = \frac{2550 |d|}{101} $$

We are given that the Mean Deviation is 255. We can set up the equation:

$$ 255 = \frac{2550 |d|}{101} $$

To find the value of d, we solve for $|d|$:

$$ |d| = \frac{255 \times 101}{2550} $$

$$ |d| = \frac{255}{2550} \times 101 $$

$$ |d| = \frac{1}{10} \times 101 $$

$$ |d| = 10.1 $$

This indicates that $d = 10.1$ or $d = -10.1$. Considering the options provided, the value of d is $10.1$.