The sum of the first n terms of an AP is 2n and the sum of the first 2n terms is 5n. Find the first term and common difference.

The sum of the first $n$ terms of an arithmetic progression (AP) is given by the formula:

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

We are provided with two specific conditions:

$$S_n = \frac{n}{2}[2a + (n-1)d] = 2n \quad (1)$$ $$S_{2n} = \frac{2n}{2}[2a + (2n-1)d] = 5n \quad (2)$$

From equation (1):

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

Assuming $n \neq 0$, we can divide both sides by $n$ and multiply by 2:

$$2a + (n-1)d = 4 \quad (3)$$

From equation (2):

$$n[2a + (2n-1)d] = 5n$$

Assuming $n \neq 0$, we can divide both sides by $n$:

$$2a + (2n-1)d = 5 \quad (4)$$

We have a system of two equations:

1. $2a + (n-1)d = 4$
2. $2a + (2n-1)d = 5$

Subtract equation (3) from equation (4):

$$(2a + (2n-1)d) - (2a + (n-1)d) = 5 - 4$$ $$2a + (2n-1)d - 2a - (n-1)d = 1$$ $$(2n-1)d - (n-1)d = 1$$

Factor out $d$:

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

Now substitute the value of $d$ into equation (3):

$$2a + (n-1)\left(\frac{1}{n}\right) = 4$$ $$2a + \frac{n-1}{n} = 4$$ $$2a = 4 - \frac{n-1}{n}$$ $$2a = \frac{4n - (n-1)}{n}$$ $$2a = \frac{4n - n + 1}{n}$$ $$2a = \frac{3n + 1}{n}$$ $$a = \frac{3n + 1}{2n}$$

The first term $a$ and the common difference $d$ are:

$$a = \boxed{\frac{3n+1}{2n}}$$ $$d = \boxed{\frac{1}{n}}$$

The formula for $S_{3n}$ is:

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

Substitute the expressions for $2a$ and $d$:

$$S_{3n} = \frac{3n}{2}\left[\left(\frac{3n+1}{n}\right) + (3n-1)\left(\frac{1}{n}\right)\right]$$ $$S_{3n} = \frac{3n}{2}\left[\frac{3n+1}{n} + \frac{3n-1}{n}\right]$$

Combine the terms inside the brackets:

$$S_{3n} = \frac{3n}{2}\left[\frac{(3n+1) + (3n-1)}{n}\right]$$ $$S_{3n} = \frac{3n}{2}\left[\frac{3n+1+3n-1}{n}\right]$$ $$S_{3n} = \frac{3n}{2}\left[\frac{6n}{n}\right]$$ $$S_{3n} = \frac{3n}{2}[6]$$ $$S_{3n} = 3n \times 3$$ $$S_{3n} = \boxed{9n}$$