Question

If the mean deviation of the data \[ 1,\ 1+d,\ 1+2d,\ \ldots,\ 1+100d \quad (d\gt 0) \] from their mean is \(255\), then \(d\) is equal to

• A. \(10.1\)
• B. \(10.2\)
• C. \(10.3\)
• D. \(10.4\)

Hint:

For an arithmetic progression with an odd number of terms, deviations from the mean occur symmetrically around zero. Use symmetry to simplify mean deviation calculations.

Answer 1

The Correct Option is A

Step 1: Identify the arithmetic progression.
The data is \[ 1,\ 1+d,\ 1+2d,\ \ldots,\ 1+100d \] This is an A.P. with: \[ a=1,\qquad \text{common difference}=d \] Number of terms: \[ n=101 \]

Step 2: Find the mean.
For an A.P., mean is \[ \frac{\text{first term}+\text{last term}}{2} \] Thus, \[ \bar x=\frac{1+(1+100d)}{2} \] \[ =\frac{2+100d}{2} \] \[ =1+50d \]

Step 3: Find deviations from the mean.
The deviations are: \[ -50d,-49d,\ldots,-d,0,d,\ldots,49d,50d \] Their absolute values are: \[ 50d,49d,\ldots,d,0,d,\ldots,49d,50d \]

Step 4: Compute mean deviation.
Mean deviation about mean is \[ \frac{\text{sum of absolute deviations}}{101} \] Now, \[ \text{sum of absolute deviations} = 2d(1+2+3+\cdots+50) \] Using \[ 1+2+\cdots+50=\frac{50\cdot51}{2} \] we get \[ \text{sum}=2d\cdot \frac{50\cdot51}{2} \] \[ =2550d \] Hence, \[ \text{Mean Deviation}=\frac{2550d}{101} \] Given this equals \(255\), \[ \frac{2550d}{101}=255 \]

Step 5: Solve for \(d\).
\[ 2550d=255\times101 \] \[ d=\frac{255\times101}{2550} \] \[ d=\frac{101}{10} \] \[ d=10.1 \]

Step 6: Final conclusion.
Therefore, \[ \boxed{10.1} \]