Arithmetic Sum C08327

1. **State the problem:** We are given an arithmetic sequence where the sum of the first 10 terms is $120$ and the sum of the first 20 terms is $840$. We need to find the sum of the first 30 terms. 2. **Recall the formula for the sum of the first $n$ terms of an arithmetic sequence:** $$S_n = \frac{n}{2} (2a_1 + (n-1)d)$$ where $S_n$ is the sum of the first $n$ terms, $a_1$ is the first term, and $d$ is the common difference. 3. **Set up equations using the given information:** - For $n=10$: $$S_{10} = \frac{10}{2} (2a_1 + 9d) = 120$$ Simplify: $$5(2a_1 + 9d) = 120$$ $$2a_1 + 9d = \frac{120}{5} = 24$$ - For $n=20$: $$S_{20} = \frac{20}{2} (2a_1 + 19d) = 840$$ Simplify: $$10(2a_1 + 19d) = 840$$ $$2a_1 + 19d = \frac{840}{10} = 84$$ 4. **Solve the system of equations:** $$\begin{cases} 2a_1 + 9d = 24 \\ 2a_1 + 19d = 84 \end{cases}$$ Subtract the first equation from the second: $$ (2a_1 + 19d) - (2a_1 + 9d) = 84 - 24 $$ $$ 10d = 60 $$ $$ d = \frac{60}{10} = 6 $$ 5. **Find $a_1$ using $d=6$ in the first equation:** $$ 2a_1 + 9(6) = 24 $$ $$ 2a_1 + 54 = 24 $$ $$ 2a_1 = 24 - 54 = -30 $$ $$ a_1 = \frac{-30}{2} = -15 $$ 6. **Find the sum of the first 30 terms $S_{30}$:** $$ S_{30} = \frac{30}{2} (2a_1 + 29d) $$ $$ = 15 (2(-15) + 29(6)) $$ $$ = 15 (-30 + 174) $$ $$ = 15 (144) $$ $$ = 2160 $$ **Final answer:** The sum of the first 30 terms is $\boxed{2160}$.