NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

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

Choose the correct answer:

Explanation

1. Find the Mean ( $\bar{x}$ )

The given numbers are in an Arithmetic Progression: $a=1$ , common difference $= d$ , and number of terms $n = 101$ (since the sequence goes from $0d$ to $100d$ ).

For an AP, the mean is the average of the first and last terms:

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

2. Calculate the Mean Deviation (M.D.)

The formula for Mean Deviation from the mean is:

M.D. = \frac{\sum_{i=0}^{100} |x_i - \bar{x}|}{n}

Substitute $x_i = 1 + id$ and $\bar{x} = 1 + 50d$ :

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

Assuming $d > 0$ , the sum becomes:

\sum_{i=0}^{100} |i - 50|d = d [ |0-50| + |1-50| + \dots + |50-50| + \dots + |100-50| ]

\text{Sum} = d [ 50 + 49 + \dots + 1 + 0 + 1 + \dots + 49 + 50 ]

This consists of two identical series of $1$ to $50$ :

\text{Sum} = 2d \left( \frac{50 \times 51}{2} \right) = 2550d

3. Solve for $d$

Given $M.D. = 255$ and $n = 101$ :

255 = \frac{2550d}{101}

Multiply by $101$ :

255 \times 101 = 2550d

Divide by $255$ :

101 = 10d

d = \frac{101}{10} = 10.1

Correct Option: (b)