Let $a$ be the first term, $d$ be the common difference, and $n$ be the number of terms in the arithmetic progression (AP). The formula for the $k$-th term of an AP is $a_k = a + (k-1)d$. The formula for the sum of $n$ terms of an AP is $S_n = \frac{n}{2}(a_1 + a_n)$, where $a_1$ is the first term and $a_n$ is the last term. Step 1: Use the given information to form equations. 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 $S_n = \frac{n}{2}(a_1 + a_n)$: $258 = \frac{n}{2}(43)$ Step 2: Find the number of terms ($n$). From $258 = \frac{n}{2}(43)$: $258 \times 2 = 43n$ $516 = 43n$ $n = \frac{516}{43}$ $n = 12$ Step 3: Use the number of terms to form another equation. 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) Step 4: Solve the system of linear equations for $a$ and $d$. 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: $2(11 - 2d) + 11d = 43$ $22 - 4d + 11d = 43$ $22 + 7d = 43$ $7d = 43 - 22$ $7d = 21$ $d = \frac{21}{7}$ $d = 3$ Step 5: Find the value of $a$. 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 $\boxed{3}$