2. The arithmetic mean of several consecutive natural numbers is 5 times greater than the smallest of them. How many times is the arithmetic mean less than the largest of these numbers?
Answer: $9 / 5 = 1.8$ times. Solution. Let the total number of numbers be $d$, and $n-$ the smallest of them. Then their arithmetic mean is

$$
(n+(n+1)+\ldots+(n+d-1)) / d=(n d+(1+2+\ldots+d-1)) / d=n+d(d-1) / 2 d=n+(d-1) / 2
$$

On the other hand, it equals $5 n$, from which $d-1=8 n$. Therefore, the largest number is $n+8 n=9 n$, which is greater than the arithmetic mean by $9 n / 5 n=9 / 5$ times.