4.070 It is known that for any $n$ the sum of the first $n$ terms of a certain numerical sequence is expressed by the formula $S_{n}=2 n^{2}+3 n$. Find the tenth term of this sequence and prove that this sequence is an arithmetic progression.
Solution. Let $a_{1}, a_{2}, \ldots a_{n}$ be the given sequence. Then

$a_{n}=S_{n}-S_{n-1}=2 n^{3}+3 n-\left(2(n-1)^{2}+3(n-1)\right)=4 n+1$ for $n \geq 2$. Further, $a_{n+1}-a_{n}=(4(n+1)+1)-(4 n+1)=4$ for $n \geq 2$. It remains to check the difference $a_{2}-a_{1}: a_{2}=4 \cdot 2+1=9, a_{1}=S_{1}=2 \cdot 1^{2}+3 \cdot 1=5 \Rightarrow a_{2}-a_{1}=4$.

Therefore, $a_{1}, a_{2}, \ldots, a_{n}$ is an arithmetic progression and $a_{10}=41$.

Answer: 41.