Let $S_{51}$ be the sum of all 51 numbers, and $A_{51}$ be their average. We know $A_{51} = 1000$. Therefore, $S_{51} = 51 \times 1000 = 51000$.
When the first number ($n_1$) is removed, there are 50 numbers remaining. The average of these 50 numbers ($A_{50}$) is $A_{51} - 1$.
The sum of these 50 numbers is $S_{50} = S_{51} - n_1$.
So, $A_{50} = \frac{S_{50}}{50} = \frac{S_{51} - n_1}{50}$.
We are given $A_{50} = A_{51} - 1 = 1000 - 1 = 999$.
Substituting the values: $999 = \frac{51000 - n_1}{50}$.
$999\times 50 = 51000 - n_1$
$\rightarrow 49950 = 51000 - n_1 $
$\rightarrow n_1 = 51000 - 49950 = 50$.
Continuing this logic for subsequent removals:
When the first two numbers ($n_1, n_2$) are removed, the average of the remaining 49 numbers ($A_{49}$) is $A_{51} - 2$.
$A_{49} = 1000 - 2 = 998$.
The sum of these 49 numbers is $S_{49} = S_{51} - n_1 - n_2$.
$A_{49} = \frac{S_{49}}{49} = \frac{S_{51} - n_1 - n_2}{49}$.
$998 = \frac{51000 - 50 - n_2}{49} $
$\rightarrow 998 \times 49 = 50950 - n_2 $
$\rightarrow 48902 = 50950 - n_2 $
$\rightarrow n_2 = 50950 - 48902 = 2048$.
This pattern continues. For each number removed from the front, the average decreases by 1, and the count of numbers decreases by 1.
