1. **Problem statement:** Find the sum of the first 40 terms of an arithmetic series given that the sum of the first 12 terms is 186 and the 20th term is 83. 2. **Recall the formulas:** - The sum of the first $n$ terms of an arithmetic series is given by: $$S_n = \frac{n}{2} (2a + (n-1)d)$$ where $a$ is the first term and $d$ is the common difference. - The $n$th term of an arithmetic series is: $$a_n = a + (n-1)d$$ 3. **Use the given information:** - Sum of first 12 terms: $$S_{12} = 186 = \frac{12}{2} (2a + 11d) = 6(2a + 11d)$$ - 20th term: $$a_{20} = a + 19d = 83$$ 4. **From the sum equation:** $$6(2a + 11d) = 186$$ Divide both sides by 6: $$\cancel{6}(2a + 11d) = \cancel{6}31$$ $$2a + 11d = 31$$ 5. **From the 20th term equation:** $$a + 19d = 83$$ 6. **Solve the system of equations:** From equation 1: $$2a + 11d = 31$$ Multiply equation 2 by 2: $$2a + 38d = 166$$ Subtract equation 1 from this: $$2a + 38d - (2a + 11d) = 166 - 31$$ $$2a + 38d - 2a - 11d = 135$$ $$27d = 135$$ $$d = \frac{135}{27} = 5$$ 7. **Find $a$:** Using $a + 19d = 83$: $$a + 19(5) = 83$$ $$a + 95 = 83$$ $$a = 83 - 95 = -12$$ 8. **Find the sum of the first 40 terms:** $$S_{40} = \frac{40}{2} (2a + 39d) = 20 (2(-12) + 39(5))$$ Calculate inside the parentheses: $$2(-12) = -24$$ $$39(5) = 195$$ $$-24 + 195 = 171$$ So, $$S_{40} = 20 \times 171 = 3420$$ **Final answer:** $$\boxed{3420}$$