4.1. Given an arithmetic progression $a_{1}, a_{2}, \ldots, a_{100}$. It is known that $a_{3}=9.5$, and the common difference of the progression $d=0.6$. Find the sum $\left\{a_{1}\right\}+\left\{a_{2}\right\}+\ldots+\left\{a_{100}\right\}$. The notation $\{x\}$ represents the fractional part of the number $x$, i.e., the difference between $x$ and the greatest integer not exceeding $x$, for example: $\{4.7\}=0.7,\left\{-5 \frac{1}{3}\right\}=\frac{2}{3}$, etc.
Answer: 50

Solution. Let's look at the first digits after the decimal point in the decimal representation of the terms of the progression. Since $d=0.6$ and $a_{3}=9.5$, the sequence of these digits is: $3,9,5,1,7,3,9,5,1,7, \ldots$ It is periodic with a period of 5. Then the sum of the fractional parts of any five consecutive terms in the progression is $0.3+0.9+0.5+0.1+0.7=2.5$. Therefore, the desired sum of the fractional parts is $20 \cdot 2.5=50$.