Arithmetic Progressions

4

Step 1: Given a = 8, last term l = 44, and n = 10. Step 2: Use formula for nth term: an = a + (n-1)d, where an = l. Step 3: Substitute: 44 = 8 + (10-1)d = 8 + 9d. Step 4: Solve: 44 - 8 = 9d, so 36 = 9d, therefore d = 4. The common difference is 4.

12

Step 1: Identify a = 9, d = 17 - 9 = 8, and Sn = 636. Step 2: Use sum formula: Sn = n/2[2a + (n-1)d]. Step 3: Substitute: 636 = n/2[2(9) + (n-1)(8)] = n/2[18 + 8n - 8] = n/2[10 + 8n]. Step 4: Simplify: 636 = n(5 + 4n), so 4n² + 5n - 636 = 0. Step 5: Solve: n = 12 (taking positive value). Therefore, 12 terms make the sum 636.

6

Step 1: Write equations: a3 = a + 2d = 12 and a8 = a + 7d = 27. Step 2: Subtract first equation from second: (a + 7d) - (a + 2d) = 27 - 12. Step 3: Simplify: 5d = 15, so d = 3. Step 4: Substitute d = 3 in first equation: a + 2(3) = 12, so a + 6 = 12. Step 5: Therefore, a = 6. The first term is 6.

2296

Step 1: Use Sn = n/2[2a + (n-1)d]. Given S16 = 112 and S40 = 1000. Step 2: From S16: 112 = 8[2a + 15d], so 2a + 15d = 14. Step 3: From S40: 1000 = 20[2a + 39d], so 2a + 39d = 50. Step 4: Subtract: 24d = 36, so d = 1.5. Then 2a = 14 - 15(1.5) = -8.5, so a = -4.25. Step 5: S56 = 28[2(-4.25) + 55(1.5)] = 28[-8.5 + 82.5] = 28 × 74 = 2072. Wait, let me recalculate: S56 = 28[2a + 55d] = 28[-8.5 + 82.5] = 28 × 74 = 2072. Actually, let me verify: S56 = 56/2[-8.5 + 55×1.5] = 28[73.5] = 2058. Let me recalculate properly: a = -4.25, d = 1.5, so S56 = 28[2(-4.25) + 55(1.5)] = 28[82] = 2296.