1. **State the problem:** We are given the 11th term ($a_{11}$) and 19th term ($a_{19}$) of an arithmetic progression (AP) as 42 and 74 respectively. We need to find the sum of the first 20 terms ($S_{20}$) of this AP. 2. **Recall the formulas:** - The $n$th term of an AP is given by $$a_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference. - The sum of the first $n$ terms of an AP is $$S_n = \frac{n}{2} [2a + (n-1)d]$$. 3. **Use the given information:** - For the 11th term: $$a_{11} = a + 10d = 42$$ - For the 19th term: $$a_{19} = a + 18d = 74$$ 4. **Find $d$ by subtracting the two equations:** $$a + 18d - (a + 10d) = 74 - 42$$ $$8d = 32$$ $$d = 4$$ 5. **Find $a$ by substituting $d$ back into one equation:** $$a + 10(4) = 42$$ $$a + 40 = 42$$ $$a = 2$$ 6. **Calculate the sum of the first 20 terms:** $$S_{20} = \frac{20}{2} [2(2) + (20-1)(4)]$$ $$= 10 [4 + 76] = 10 \times 80 = 800$$ **Final answer:** The sum of the first 20 terms of the AP is $800$.