4.041. Find the positive integer $n$ from the equation

$$
(3+6+9+\ldots+3(n-1))+\left(4+5.5+7+\ldots+\frac{8+3 n}{2}\right)=137
$$
## Solution.

In the first parentheses, there is the sum of the terms of an arithmetic progression $S_{k}$ where $a_{1}=3, d=3, a_{k}=3(n-1), k=\frac{a_{k}-a_{1}}{d}+1=\frac{3 n-3-3}{3}+1=n-1$; in the second parentheses, there is the sum of the terms of an arithmetic progression where $b_{1}=4, d=1.5, a_{m}=\frac{8+3 n}{2}, m=\frac{a_{m}-a_{1}}{d}+1=\frac{\frac{8+3 n}{2}-4}{1.5}+1=n+1$.

Then the original equation takes the form

$$
\frac{3+3(n-1)}{2} \cdot(n-1)+\frac{4+\frac{8+3 n}{2}}{2} \cdot(n+1)=137 \Leftrightarrow 9 n^{2}+13 n-532=0
$$

From this, $n_{1}=-\frac{76}{9}, n_{2}=7 ; n_{1}=-\frac{76}{9}$ is not suitable, as $n$ is an integer.

Answer: 7.