We are given an arithmetic sequence $\{a_n\}$ with the sum of the first $n$ terms denoted by $S_n$. We are given that $a_5 = 13$ and $S_5 = 35$. We are asked to find: (1) The general term formula $a_n$ of the sequence $\{a_n\}$. (2) The sum of the first $n$ terms $S_n$ of the sequence $\{a_n\}$.

1. Problem Description

We are given an arithmetic sequence $\{a_n\}$ with the sum of the first $n$ terms denoted by $S_n$. We are given that $a_5 = 13$ and $S_5 = 35$. We are asked to find:

(1) The general term formula $a_n$ of the sequence $\{a_n\}$.

(2) The sum of the first $n$ terms $S_n$ of the sequence $\{a_n\}$.

2. Solution Steps

(1) Since $\{a_n\}$ is an arithmetic sequence, we have:

$a_n = a_1 + (n-1)d$

$S_n = \frac{n(a_1 + a_n)}{2} = \frac{n[2a_1 + (n-1)d]}{2}$

We are given $a_5 = 13$, so

$a_5 = a_1 + 4d = 13$ (1)

We are also given $S_5 = 35$, so

$S_5 = \frac{5(2a_1 + 4d)}{2} = 35$

$5(2a_1 + 4d) = 70$

$2a_1 + 4d = 14$

$a_1 + 2d = 7$ (2)

Subtracting equation (2) from equation (1), we get:

$(a_1 + 4d) - (a_1 + 2d) = 13 - 7$

$2d = 6$

$d = 3$

Substituting $d=3$ into equation (2), we get:

$a_1 + 2(3) = 7$

$a_1 + 6 = 7$

$a_1 = 1$

Therefore, the general term formula is:

$a_n = a_1 + (n-1)d = 1 + (n-1)3 = 1 + 3n - 3 = 3n - 2$

(2) The sum of the first $n$ terms is:

$S_n = \frac{n[2a_1 + (n-1)d]}{2} = \frac{n[2(1) + (n-1)3]}{2} = \frac{n[2 + 3n - 3]}{2} = \frac{n(3n - 1)}{2}$

$S_n = \frac{3n^2 - n}{2}$

3. Final Answer

(1) $a_n = 3n - 2$

(2) $S_n = \frac{3n^2 - n}{2}$