(1) For the arithmetic sequence $\left\{a_{n}\right\}$ with the general term formula $a_{n}=3 n-2$, find the formula for the sum of the first $n$ terms.
(2) For the arithmetic sequence $\left\{a_{n}\right\}$ with the sum of the first $n$ terms given by the formula $S_{n}=5 n^{2}+3 n$, find its first 3 terms, and determine its general term formula.
Solving (1) yields $S_{n}=\frac{3}{2} n^{2}-\frac{1}{2} n$. The general term formula of this arithmetic sequence is a linear function of $n$, and the sum of the first $n$ terms is a quadratic function of $n$. Solving (2) yields $a_{n}=S_{n}-S_{n-1}$ $=\left(5 n^{2}+3 n\right)-\left[5(n-1)^{2}+3(n-1)\right]$ $=10 n-2 \quad(n \geqslant 2)$. When $n=1$, $10 \times 1-2=8, S_{1}=5 \times 1^{2}+3 \times 1=8$. Therefore, $a_{n}=10 n-2$ also applies to $n=1$. From this problem, we can also see that the general term formula of an arithmetic sequence is a linear function of $n$, and the formula for the sum of the first $n$ terms is a quadratic function of $n$. Can this be generalized to the following proposition: A sequence is an arithmetic sequence if and only if its general term formula is a linear function of the term number $n$. A sequence is an arithmetic sequence if and only if the formula for the sum of its first $n$ terms is a quadratic function of the term number $n$.

If the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence, then $a_{n}=a_{1}+$ $(n-1) d=n d+(x, \cdots d)$. When $d \neq 0$, it is a linear function of $n$. When $a=0$, it is a zero-degree function of $n$, i.e., a constant. Conversely, if the general term formula of the sequence $\left\{a_{n}\right\}$ is $a_{n}=A n+B$, then $a_{n}-a_{n-1}=(A n+B)-$ $[A(n-1)+B]=A, A$ is a constant, so the sequence is an arithmetic sequence. When $A \neq 0$, $a_{n}=A n+B$ is a linear function of $n$, at this time the sequence is an arithmetic sequence with a non-zero common difference. When $A=0$, $a_{n}=A n+B=B$ is a zero-degree function of $n$, at this time, the sequence is an arithmetic sequence with a common difference of zero, i.e., a constant sequence.

From the above, we can conclude that a sequence is an arithmetic sequence if and only if its general term formula is $a_{n}=A n+B$, where $A$ and $B$ are constants.

Next, we study the second proposition. If the sequence $\left\{a_{n}\right\}$ is an arithmetic sequence, then the sum of its first $n$ terms is
$$
S_{n}=\frac{n\left[2 a_{1}+(n-1) d\right]}{2}=\frac{d}{2} n^{2}+\frac{2 a_{1}-d}{2} n .
$$

It is in the form of $A n^{2}+B n$, when $d \neq 0$, it is a quadratic function of $n$; when $d=0, a_{1} \neq 0$, it is $a_{1} n$, a linear function of $n$; when $d=0, a_{1}=0, S_{n}=0$, it is a zero-degree function of $n$. Conversely, if the formula for the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is $S_{n}=A n^{2}+B n+C$, then $a_{n}=: S_{n}-S_{n-1}$ $=\left(A n^{2}+B n+C\right)-\left[A(n-1)^{2}+B(n-1)\right.$ $+C]=2 A \tilde{n}+(B-\hat{A}) \quad(n \geqslant 2), a_{n}-a_{n-1}$ $=[2 A n+(B-A)]-[2 A(n-1)+(B-$ A) $]=2 A$. From this, we can see that from the second term onwards, this sequence is an arithmetic sequence, $a_{1}=S_{1}=A+B+C, a_{1}=$ $2 A-1+(B-A)=A+B$. To make $a_{n}=2 A n$ + $(B-A)$ also apply to $a_{1}$, we should let $C=0$. That is, when $C=$ 0, this sequence is an arithmetic sequence from the first term onwards.

From this, we can conclude that a sequence is an arithmetic sequence if and only if the sum of its first $n$ terms is $S_{n}=A n^{2}+B n$ (where $A$ and $B$ are constants).