Let a be the first term, d be the common difference, and n be the number of terms in the arithmetic progression (AP).

We are given:

1. The sum of the first and last term is 43. $a_1 + a_n = 43$ Since $a_1 = a$, this is $a + a_n = 43$.

2. The third term is 11. $a_3 = a + (3-1)d = 11$ $a + 2d = 11$ (Equation 1)

3. The sum of all terms is 258. $S_n = 258$ Using the sum formula $S_n = \frac{n}{2}(a_1 + a_n)$: $258 = \frac{n}{2}(43)$

From $258 = \frac{n}{2}(43)$:
$258 \times 2 = 43n$
$516 = 43n$
$n = \frac{516}{43}$
$n = 12$

Since $n=12$, the last term is $a_{12}$.
We know $a + a_n = 43$, so $a + a_{12} = 43$.
The formula for $a_{12}$ is $a + (12-1)d = a + 11d$.
Substitute this into the sum of first and last term:
$a + (a + 11d) = 43$
$2a + 11d = 43$ (Equation 2)

We have two equations:

1. $a + 2d = 11$
2. $2a + 11d = 43$

From Equation 1, express $a$ in terms of $d$:
$a = 11 - 2d$

Substitute this expression for $a$ into Equation 2:
$2(11 - 2d) + 11d = 43$
$22 - 4d + 11d = 43$
$22 + 7d = 43$
$7d = 43 - 22$
$7d = 21$
$d = \frac{21}{7}$
$d = 3$

Substitute $d=3$ back into the expression for $a$:
$a = 11 - 2(3)$
$a = 11 - 6$
$a = 5$

The first term is 5 and the common difference is 3.

The first term is $\boxed{5}$
The common difference is $\boxed{3}$