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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Let's break this problem down into two parts. a) Find the first term and the common difference Let the first term be a and the common difference be d. The formula for the n-th term of an arithmetic progression is aₙ = a + (n-1)d. Step 1: Use the first piece of information to form an equation. "The sum of the sixth and eighth terms of an arithmetic progression is 40." The sixth term is a₆ = a + (6-1)d = a + 5d. The eighth term is a₈ = a + (8-1)d = a + 7d. So, a₆ + a₈ = 40 (a + 5d) + (a + 7d) = 40 2a + 12d = 40 Divide by 2: a + 6d = 20 (Equation 1) Step 2: Use the second piece of information to form another equation. "The seventh term of this progression is 4 times the second term." The seventh term is a₇ = a + (7-1)d = a + 6d. The second term is a₂ = a + (2-1)d = a + d. So, a₇ = 4 × a₂ a + 6d = 4(a + d) a + 6d = 4a + 4d Rearrange the terms to group 'a' and 'd': 6d - 4d = 4a - a 2d = 3a (Equation 2) Step 3: Solve the system of equations to find a and d. From Equation 2, we can express a in terms of d: a = (2/3)d Substitute this expression for a into Equation 1: (2/3)d + 6d = 20 To add the 'd' terms, find a common denominator: (2/3)d + (18/3)d = 20 (20/3)d = 20 Multiply both sides by (3/20): d = 20 × (3/20) d = 3 Now substitute the value of d back into Equation 2 to find a: 3a = 2(3) 3a = 6 a = 6 / 3 a = 2 The first term is 2 and the common difference is 3. b) The least number of terms required for the sum of the progression to exceed 1000 Step 1: Write the formula for the sum of an arithmetic progression. The sum of the first n terms is Sₙ = (n/2)[2a + (n-1)d]. We need Sₙ > 1000. Substitute a = 2 and d = 3 into the formula: (n/2)[2(2) + (n-1)3] > 1000 (n/2)[4 + 3n - 3] > 1000 (n/2)[3n + 1] > 1000 Step 2: Solve the inequality for n. Multiply both sides by 2: n(3n + 1) > 2000 3n² + n > 2000 3n² + n - 2000 > 0 To find the values of n that satisfy this inequality, we first find the roots of the quadratic equation 3n² + n - 2000 = 0 using the quadratic formula: n = [-b ± √(b² - 4ac)] / (2a) Here, a=3, b=1, c=-2000. n = [-1 ± √(1² - 4 × 3 × -2000)] / (2 × 3) n = [-1 ± √(1 + 24000)] / 6 n = [-1 ± √(24001)] / 6 Calculate the square root: √(24001) ≈ 154.9225 Now find the two possible values for n: n₁ = (-1 + 154.9225) / 6 = 153.9225 / 6 ≈ 25.65 n₂ = (-1 - 154.9225) / 6 = -155.9225 / 6 ≈ -25.99 Since n represents the number of terms, it must be a positive integer. The inequality 3n² + n - 2000 > 0 holds for n values greater than the positive root. So, n > 25.65. Step 3: Determine the least integer value for n. Since n must be an integer and greater than 25.65, the smallest integer value for n is 26. Let's check: For n = 25: S₂₅ = (25/2)[2(2) + (25-1)3] = (25/2)[4 + 24 × 3] = (25/2)[4 + 72] = (25/2)[76] = 25 × 38 = 950. (This is not > 1000) For n = 26: S₂₆ = (26/2)[2(2) + (26-1)3] = 13[4 + 25 × 3] = 13[4 + 75] = 13[79] = 1027. (This is > 1000) The least number of terms required for the sum of the progression to exceed 1000 is 26. What's next?