Step 1: Define the terms for an arithmetic progression. Let $a$ be the first term and $d$ be the common difference. The $n$-th term is given by $T_n = a + (n-1)d$. The sum of the first $n$ terms is given by $S_n = \frac{n}{2}(2a + (n-1)d)$. Step 2: Formulate equations from the given information. From the statement "The eleventh term of an arithmetic progression is four times its second term": $$T_{11} = 4 \times T_2$$ $$a + (11-1)d = 4(a + (2-1)d)$$ $$a + 10d = 4(a + d) \quad (1)$$ From the statement "The sum of the first seven terms of the same progression is 175": $$S_7 = 175$$ $$\frac{7}{2}(2a + (7-1)d) = 175$$ $$\frac{7}{2}(2a + 6d) = 175 \quad (2)$$ Step 3: Simplify Equation (1). $$a + 10d = 4a + 4d$$ $$10d - 4d = 4a - a$$ $$6d = 3a$$ $$a = 2d \quad (3)$$ Step 4: Simplify Equation (2). Multiply both sides by $\frac{2}{7}$: $$2a + 6d = 175 \times \frac{2}{7}$$ $$2a + 6d = 25 \times 2$$ $$2a + 6d = 50$$ Divide by 2: $$a + 3d = 25 \quad (4)$$ Step 5: Substitute Equation (3) into Equation (4) to solve for $d$. Substitute $a = 2d$ into $a + 3d = 25$: $$2d + 3d = 25$$ $$5d = 25$$ $$d = \frac{25}{5}$$ $$d = 5$$ Step 6: Substitute the value of $d$ back into Equation (3) to solve for $a$. $$a = 2d$$ $$a = 2 \times 5$$ $$a = 10$$ The first term is $10$ and the common difference is $5$. The final answer is $\boxed{\text{First term} = 10, \text{Common difference} = 5}$.