This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions for the remaining parts of question 1 and for question 2. e) Step 1: Find the common difference. The terms are -3, 6, 15, 24. The difference between consecutive terms is 6 - (-3) = 9, 15 - 6 = 9, 24 - 15 = 9. The common difference is 9. Step 2: Determine the n^th term. Since the common difference is 9, the n^th term will be of the form 9n + c. For n=1, the term is -3. So, 9(1) + c = -3 9 + c = -3 c = -12. Thus, the n^th term is 9n - 12. Step 3: Calculate the 10^th term. Substitute n=10 into the n^th term formula: T_10 = 9(10) - 12 = 90 - 12 = 78. The n^th term is: 9n - 12 The 10^th term is: 78 f) Step 1: Find the common difference. The terms are 13, 17, 21, 25. The difference between consecutive terms is 17 - 13 = 4, 21 - 17 = 4, 25 - 21 = 4. The common difference is 4. Step 2: Determine the n^th term. Since the common difference is 4, the n^th term will be of the form 4n + c. For n=1, the term is 13. So, 4(1) + c = 13 4 + c = 13 c = 9. Thus, the n^th term is 4n + 9. Step 3: Calculate the 10^th term. Substitute n=10 into the n^th term formula: T_10 = 4(10) + 9 = 40 + 9 = 49. The n^th term is: 4n + 9 The 10^th term is: 49 g) Step 1: Find the common difference. The terms are -6, 10, 26, 42. The difference between consecutive terms is 10 - (-6) = 16, 26 - 10 = 16, 42 - 26 = 16. The common difference is 16. Step 2: Determine the n^th term. Since the common difference is 16, the n^th term will be of the form 16n + c. For n=1, the term is -6. So, 16(1) + c = -6 16 + c = -6 c = -22. Thus, the n^th term is 16n - 22. Step 3: Calculate the 10^th term. Substitute n=10 into the n^th term formula: T_10 = 16(10) - 22 = 160 - 22 = 138. The n^th term is: 16n - 22 The 10^th term is: 138 Problem solving - First sequence The given terms are T_3 = 13, T_6 = 40, T_9 = 85, T_12 = 148. Step 1: Determine the type of sequence. Let the n^th term be T_n = an^2 + bn + c. Using the given terms: For n=3: 9a + 3b + c = 13 (Eq 1) For n=6: 36a + 6b + c = 40 (Eq 2) For n=9: 81a + 9b + c = 85 (Eq 3) Step 2: Solve the system of equations. Subtract (Eq 1) from (Eq 2): 27a + 3b = 27 (Eq 4) Subtract (Eq 2) from (Eq 3): 45a + 3b = 45 (Eq 5) Subtract (Eq 4) from (Eq 5): 18a = 18 a = 1. Substitute a=1 into (Eq 4): 27(1) + 3b = 27 3b = 0 b = 0. Substitute a=1, b=0 into (Eq 1): 9(1) + 3(0) + c = 13 9 + c = 13 c = 4. Step 3: Determine the n^th term. The n^th term is T_n = 1n^2 + 0n + 4 = n^2 + 4. Step 4: Calculate the 10^th term. Substitute n=10 into the n^th term formula: T_10 = 10^2 + 4 = 100 + 4 = 104. The n^th term is: n^2 + 4 The 10^th term is: 104 Problem solving - Second sequence The given terms are T_6 = 199, T_12 = 1711, T_18 = 5815. We need to find T_10. Step 1: Determine the type of sequence. Let the n^th term be T_n = an^2 + bn + c. Using the given terms: For n=6: 36a + 6b + c = 199 (Eq 1) For n=12: 144a + 12b + c = 1711 (Eq 2) For n=18: 324a + 18b + c = 5815 (Eq 3) Step 2: Solve the system of equations. Subtract (Eq 1) from (Eq 2): 108a + 6b = 1512 (Eq 4) Subtract (Eq 2) from (Eq 3): 180a + 6b = 4104 (Eq 5) Subtract (Eq 4) from (Eq 5): 72a = 2592 a = 36. Substitute a=36 into (Eq 4): 108(36) + 6b = 1512 3888 + 6b = 1512 6b = -2376 b = -396. Substitute a=36, b=-396 into (Eq 1): 36(36) + 6(-396) + c = 199 1296 - 2376 + c = 199 -1080 + c = 199 c = 1279. Step 3: Determine the n^th term. The n^th term is T_n = 36n^2 - 396n + 1279. Step 4: Calculate the 10^th term. Substitute n=10 into the n^th term formula: T_10 = 36(10^2) - 396(10) + 1279 T_10 = 36(100) - 3960 + 1279 T_10 = 3600 - 3960 + 1279 T_10 = -360 + 1279 = 919. The n^th term is: 36n^2 - 396n + 1279 The 10^th term is: 919 2. Make notes on how you solved the sequences. For arithmetic sequences (like parts a-g), I found the common difference (d) between consecutive terms. The n^th term was then determined using the formula T_n = a + (n-1)d, where a is the first term. For quadratic sequences (like the "Problem solving" sections), I used the general form T_n = an^2 + bn + c. By substituting the given term positions (n) and their values (T_n), I created a system of simultaneous equations to solve for the coefficients a, b, and c. That's 2 down. 3 left today — send the next one.