Question

If the mean deviation of the numbers $1, 1+d, 1+2d, \dots, 1+100d$ from their mean is 255, then $d =$

• A. 10.1
• B. 10
• C. 5.05
• D. 5.1

Hint:

The mean deviation of an arithmetic progression of $(2n+1)$ terms with common difference $d$ is $\frac{n(n+1)}{2n+1}|d|$. Here, $n=50$, so $\frac{50 \times 51}{101}|d| = 255 \implies |d| = 10.1$.

Answer 1

The Correct Option is A

Step 1: Concept
The mean deviation of $N$ observations from their mean is given by $\text{M.D.} = \frac{\sum |x_i - \bar{x}|}{N}$.

Step 2: Meaning
There are $N = 101$ terms in the given arithmetic progression. Since the number of terms is odd, the mean is the middle term, which is the 51st term: $\bar{x} = 1 + 50d$.

Step 3: Analysis
The deviations of the terms from the mean $\bar{x} = 1+50d$ are: \[ |x_i - \bar{x}| = \{ 50|d|, 49|d|, \dots, 1|d|, 0, 1|d|, \dots, 50|d| \} \] Sum of deviations: \[ \sum |x_i - \bar{x}| = 2 \cdot |d| \cdot (1 + 2 + \dots + 50) = 2|d| \cdot \frac{50 \times 51}{2} = 2550|d| \] Mean deviation: \[ \text{M.D.} = \frac{2550|d|}{101} = 255 \] \[ \implies \frac{10|d|}{101} = 1 \implies |d| = \frac{101}{10} = 10.1 \]

Step 4: Conclusion
The value of the common difference $d$ is 10.1.

Final Answer: (A)