Calculate: \( \frac{4+6+8 \ldots+806}{5+10+15+\ldots+625} \) If the 7 th term of an arithmetic series is -3 and the sum of the first 16 terms is 24 , determine the first term. The first term of an arithmetic series is 5 and the last term is 61 . The of all the terms is 957 . Calculate the common difference. Tebogo wants to save money for a deposit of R51 200 on a new car. He starts saving R500

Sure, let's address each of your problems step-by-step. --- ### **1. Calculate:** \[ \frac{4 + 6 + 8 + \ldots + 806}{5 + 10 + 15 + \ldots + 625} \] **Solution:** We have two arithmetic series here: **Series A:** \(4, 6, 8, \ldots, 806\) - **First term (a₁):** 4 - **Common difference (d):** 2 (since \(6 - 4 = 2\)) - **Last term (aₙ):** 806 **Number of terms (n₁):** \[ a_n = a_1 + (n - 1)d \\ 806 = 4 + (n_1 - 1) \times 2 \\ 806 - 4 = 2(n_1 - 1) \\ 802 = 2(n_1 - 1) \\ n_1 - 1 = 401 \\ n_1 = 402 \] **Sum of Series A (S₁):** \[ S_1 = \frac{n}{2} \times (a_1 + a_n) \\ S_1 = \frac{402}{2} \times (4 + 806) \\ S_1 = 201 \times 810 \\ S_1 = 162,810 \] --- **Series B:** \(5, 10, 15, \ldots, 625\) - **First term (a₁):** 5 - **Common difference (d):** 5 - **Last term (aₙ):** 625 **Number of terms (n₂):** \[ a_n = a_1 + (n - 1)d \\ 625 = 5 + (n_2 - 1) \times 5 \\ 625 - 5 = 5(n_2 - 1) \\ 620 = 5(n_2 - 1) \\ n_2 - 1 = 124 \\ n_2 = 125 \] **Sum of Series B (S₂):** \[ S_2 = \frac{n}{2} \times (a_1 + a_n) \\ S_2 = \frac{125}{2} \times (5 + 625) \\ S_2 = 62.5 \times 630 \\ S_2 = 39,375 \] --- **Final Calculation:** \[ \frac{S_1}{S_2} = \frac{162,810}{39,375} = \frac{3618}{875} \approx 4.135 \] **Answer:** The ratio is \(\frac{3618}{875}\) or approximately **4.135**. --- ### **2. Determine the First Term of an Arithmetic Series** **Given:** - The 7th term (\(a_7\)) is -3. - The sum of the first 16 terms (\(S_{16}\)) is 24. **Solution:** **Formula for the nth term:** \[ a_n = a_1 + (n - 1)d \] For the 7th term: \[ a_7 = a_1 + 6d = -3 \quad \text{(1)} \] **Formula for the sum of the first n terms:** \[ S_n = \frac{n}{2} \times (2a_1 + (n - 1)d) \] For the first 16 terms: \[ S_{16} = \frac{16}{2} \times (2a_1 + 15d) = 24 \\ 8 \times (2a_1 + 15d) = 24 \\ 2a_1 + 15d = 3 \quad \text{(2)} \] **Solving Equations (1) and (2):** From equation (1): \[ a_1 = -3 - 6d \] Substitute \(a_1\) into equation (2): \[ 2(-3 - 6d) + 15d = 3 \\ -6 - 12d + 15d = 3 \\ -6 + 3d = 3 \\ 3d = 9 \\ d = 3 \] Now, find \(a_1\): \[ a_1 = -3 - 6(3) = -3 - 18 = -21 \] **Answer:** The first term is **-21**. --- ### **3. Calculate the Common Difference of an Arithmetic Series** **Given:** - First term (\(a_1\)) = 5 - Last term (\(a_n\)) = 61 - Sum of all terms (\(S_n\)) = 957 **Solution:** **Sum formula:** \[ S_n = \frac{n}{2} \times (a_1 + a_n) \] \[ 957 = \frac{n}{2} \times (5 + 61) \\ 957 = \frac{n}{2} \times 66 \\ 957 = 33n \\ n = \frac{957}{33} = 29 \] **Now, use the nth term formula:** \[ a_n = a_1 + (n - 1)d \\ 61 = 5 + (29 - 1)d \\ 61 = 5 + 28d \\ 61 - 5 = 28d \\ 56 = 28d \\ d = 2 \] **Answer:** The common difference is **2**. --- ### **4. Tebogo's Savings Plan** **Given:** - Tebogo wants to save **R51,200** for a car deposit. - He starts saving **R500**... *It seems the problem is incomplete. Please provide the remaining information so I can assist you further.*