4.024. The first term of an arithmetic progression is 429, and its difference is -22. How many terms of this progression need to be taken so that their sum is equal to 3069?
## Solution.

From the condition, we have $a_{1}=429, d=-22, S_{n}=3069$.

Using formula (4.5), we get

$$
\begin{aligned}
& \frac{2 \cdot 429-22(n-1)}{2} \cdot n=3069, (429-11(n-1)) n=3069 \\
& n^{2}-40 n+279=0 \Rightarrow n_{1}=9, n_{2}=31
\end{aligned}
$$

Answer: 9 or 31.