1. **Problem statement:** We have an arithmetic series where the sum of the first 5 terms is 70 and the sum of the first 11 terms is 352. We need to find the first term $a$ and common difference $d$, then find sums of various numbers of terms, and finally find the sum of terms 35 to 40. 2. **Formula for sum of first $n$ terms of an arithmetic series:** $$S_n = \frac{n}{2} [2a + (n-1)d]$$ where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, and $d$ is the common difference. 3. **Using given sums:** - For $n=5$, $S_5 = 70$: $$70 = \frac{5}{2} [2a + (5-1)d] = \frac{5}{2} (2a + 4d)$$ Multiply both sides by 2: $$140 = 5(2a + 4d)$$ Divide both sides by 5: $$\cancel{5} \times (2a + 4d) = \frac{140}{\cancel{5}}$$ $$2a + 4d = 28$$ - For $n=11$, $S_{11} = 352$: $$352 = \frac{11}{2} [2a + (11-1)d] = \frac{11}{2} (2a + 10d)$$ Multiply both sides by 2: $$704 = 11(2a + 10d)$$ Divide both sides by 11: $$\cancel{11} \times (2a + 10d) = \frac{704}{\cancel{11}}$$ $$2a + 10d = 64$$ 4. **Solve the system:** From the two equations: $$\begin{cases} 2a + 4d = 28 \\ 2a + 10d = 64 \end{cases}$$ Subtract the first from the second: $$ (2a + 10d) - (2a + 4d) = 64 - 28 $$ $$ 2a - 2a + 10d - 4d = 36 $$ $$ 6d = 36 $$ $$ d = \frac{36}{6} = 6 $$ Substitute $d=6$ into $2a + 4d = 28$: $$ 2a + 4(6) = 28 $$ $$ 2a + 24 = 28 $$ $$ 2a = 28 - 24 = 4 $$ $$ a = \frac{4}{2} = 2 $$ 5. **Find sums for other terms:** - For $n=30$: $$ S_{30} = \frac{30}{2} [2(2) + (30-1)(6)] = 15 [4 + 29 \times 6] = 15 [4 + 174] = 15 \times 178 = 2670 $$ - For $n=50$: $$ S_{50} = \frac{50}{2} [2(2) + (50-1)(6)] = 25 [4 + 49 \times 6] = 25 [4 + 294] = 25 \times 298 = 7450 $$ - For $n=100$: $$ S_{100} = \frac{100}{2} [2(2) + (100-1)(6)] = 50 [4 + 99 \times 6] = 50 [4 + 594] = 50 \times 598 = 29900 $$ 6. **Sum of terms 35 to 40:** The sum of terms from $m$ to $n$ is: $$ S_{m \text{ to } n} = S_n - S_{m-1} $$ where $S_n$ is sum of first $n$ terms. 7. **Calculate $S_{35}$ and $S_{34}$:** $$ S_{35} = \frac{35}{2} [2(2) + (35-1)(6)] = \frac{35}{2} [4 + 34 \times 6] = \frac{35}{2} [4 + 204] = \frac{35}{2} \times 208 = 35 \times 104 = 3640 $$ $$ S_{34} = \frac{34}{2} [2(2) + (34-1)(6)] = 17 [4 + 33 \times 6] = 17 [4 + 198] = 17 \times 202 = 3434 $$ 8. **Sum of terms 35 to 40:** $$ S_{35 \text{ to } 40} = S_{40} - S_{34} $$ Calculate $S_{40}$: $$ S_{40} = \frac{40}{2} [2(2) + (40-1)(6)] = 20 [4 + 39 \times 6] = 20 [4 + 234] = 20 \times 238 = 4760 $$ Therefore: $$ S_{35 \text{ to } 40} = 4760 - 3434 = 1326 $$ **Final answers:** - $a = 2$, $d = 6$ - $S_{30} = 2670$ - $S_{50} = 7450$ - $S_{100} = 29900$ - Sum of terms 35 to 40 is 1326