JEE 2023 — Mathematics PYQ

Solution

• Numbers: $a,a+1,\dots ,a+99$

• Mean ($\bar{x}$): $\frac{a+(a+99)}{2}=a+49.5$

• Mean Deviation (M.D.):

  $$M.D.=\frac{1}{100}\sum _{i=1}^{100}|x_i-(a+49.5)|$$
  $$M.D.=\frac{2}{100}\sum _{i=1}^{50}(i-0.5)$$
  $$M.D.=\frac{2}{100}\left[\frac{50}{2}(0.5+49.5)\right]=\frac{1}{50}\times 25\times 50=25$$

Kyunki $M.D.$ ki value $a_1$ par nirbhar (depend) nahi karti, isliye $a_1$ koi bhi natural number ho sakta hai.

Sahi Option: (1)