This problem involves calculating the mean deviation of an arithmetic progression (AP) and finding the value of the common difference, d. The given series is 1, 1 + d, 1 + 2d, ..., 1 + 100d. We are given that the mean deviation from the mean is 255.

Understanding the Arithmetic Progression

The sequence provided is an arithmetic progression:

• First term, $a = 1$
• Common difference = $d$
• Number of terms, $n = 101$ (from $1 + 0d$ to $1 + 100d$)
• The terms can be represented as $x_i = 1 + (i-1)d$ for $i = 1, 2, ..., 101$.

Calculating the Mean of the Series

The mean ($\bar{x}$) of an arithmetic progression can be calculated as the average of the first and the last term:

Last term, $x_{101} = 1 + (101-1)d = 1 + 100d$.

Mean, $\bar{x} = \frac{\text{First term} + \text{Last term}}{2} = \frac{1 + (1 + 100d)}{2} = \frac{2 + 100d}{2} = 1 + 50d$.

Alternatively, using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d']$: Sum, $S_{101} = \frac{101}{2}[2(1) + (101-1)d] = \frac{101}{2}[2 + 100d] = 101(1 + 50d)$. Mean, $\bar{x} = \frac{S_{101}}{n} = \frac{101(1 + 50d)}{101} = 1 + 50d$.

Calculating the Mean Deviation

The Mean Deviation (MD) is defined as the average of the absolute differences between each term and the mean:

$$ \text{MD} = \frac{1}{n} \sum_{i=1}^{n} |x_i - \bar{x}| $$

First, let's find the deviation of each term from the mean:

$x_i - \bar{x} = (1 + (i-1)d) - (1 + 50d) = (i-1)d - 50d = (i - 1 - 50)d = (i - 51)d$.

Now, we need the absolute deviations:

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

The sum of the absolute deviations is:

$$ \sum_{i=1}^{101} |x_i - \bar{x}| = \sum_{i=1}^{101} |d| \times |i - 51| = |d| \sum_{i=1}^{101} |i - 51| $$

Let's evaluate the sum $\sum_{i=1}^{101} |i - 51|$:

• When $i < 51$, $i-51$ is negative. Example: $i=1 \implies |1-51|=50$.
• When $i = 51$, $i-51$ is zero. Example: $i=51 \implies |51-51|=0$.
• When $i > 51$, $i-51$ is positive. Example: $i=52 \implies |52-51|=1$.

The sum is $ |1-51| + |2-51| + ... + |50-51| + |51-51| + |52-51| + ... + |101-51| $

$= |-50| + |-49| + ... + |-1| + |0| + |1| + ... + |50|$

$= 50 + 49 + ... + 1 + 0 + 1 + ... + 49 + 50$

This sum can be calculated as $2 \times (1 + 2 + ... + 50)$.

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

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

So, the sum $\sum_{i=1}^{101} |i - 51| = 2 \times 1275 = 2550$.

Therefore, the sum of absolute deviations is $|d| \times 2550$.

The Mean Deviation is:

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

Solving for 'd'

We are given that the Mean Deviation (MD) is 255.

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

Now, we solve for $|d|$:

$|d| \times 2550 = 255 \times 101$

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

$|d| = \frac{255 \times 101}{255 \times 10}$

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

$|d| = 10.1$

This means $d = 10.1$ or $d = -10.1$. Since 10.1 is one of the options, this is the likely value.

Conclusion

The value of $d$ for the given arithmetic progression, where the mean deviation from the mean is 255, is 10.1.