[ Arithmetic operations. Numerical identities ]

In the competition, 50 shooters participated. The first scored 60 points; the second - 80; the third - the arithmetic mean of the points of the first two; the fourth - the arithmetic mean of the points of the first three. Each subsequent shooter scored the arithmetic mean of the points of all previous shooters. How many points did the 42nd shooter score? And the 50th?
Do you remember that if you add a number equal to the arithmetic mean of a group of numbers to that group, the arithmetic mean of the new group will be equal to the arithmetic mean of the initial group.

## Solution

The third player scored $(60+80): 2=70$ points. Each subsequent player also scored 70 points: if you add a number equal to the arithmetic mean of a group of numbers to that group, the arithmetic mean of the new group will be equal to the arithmetic mean of the initial group.