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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Another one Kwanga — let's solve it. Question 2 2.1.1 The common ratio. Step 1: Identify the given values for the geometric sequence. The first term is a = (3)/(128). The sixth term is T_6 = (3)/(4). The formula for the n-th term of a geometric sequence is T_n = a · r^n-1. Step 2: Substitute the given values into the formula for T_6. T_6 = a · r^6-1 (3)/(4) = (3)/(128) · r^5 Step 3: Solve for r. Divide both sides by (3)/(128): (3)/(4) ÷ (3)/(128) = r^5 (3)/(4) × (128)/(3) = r^5 (128)/(4) = r^5 32 = r^5 Take the fifth root of both sides: r = [5]32 r = 2 The common ratio is 2. 2.1.2 The sum of the first 20 terms, correct to the nearest integer. Step 1: Identify the values needed for the sum formula. First term a = (3)/(128). Common ratio r = 2 (from 2.1.1). Number of terms n = 20. The formula for the sum of the first n terms of a geometric sequence is S_n = (a(r^n - 1))/(r - 1). Step 2: Substitute the values into the sum formula. S_20 = (3)/(128)(2^20 - 1)2 - 1 S_20 = (3)/(128)(2^20 - 1) Step 3: Calculate 2^20. 2^20 = 1,048,576 Step 4: Substitute the value and calculate S_20. S_20 = (3)/(128)(1,048,576 - 1) S_20 = (3)/(128)(1,048,575) S_20 = (3,145,725)/(128) S_20 ≈ 24,575.9765625 Step 5: Round the sum to the nearest integer. S_20 ≈ 24,576 The sum of the first 20 terms is 24,576. 2.2 Prove that a = sqrt(3)(sqrt(5)-1)2. Step 1: Use the property of a geometric sequence. Given that a, b, sqrt(3) are three consecutive terms of a geometric sequence, the common ratio is constant. (b)/(a) = sqrt(3)b b^2 = asqrt(3) (*) Step 2: Use the property of a right-angled triangle. Given that a, b, sqrt(3) are the sides of a right-angled triangle, and a < b < sqrt(3), then sqrt(3) must be the hypotenuse. By the Pythagorean theorem: a^2 + b^2 = (sqrt(3))^2 a^2 + b^2 = 3 (**) Step 3: Substitute equation () into equation (*). a^2 + asqrt(3) = 3 Rearrange into a quadratic equation: a^2 + asqrt(3) - 3 = 0 Step 4: Solve the quadratic equation for a using the quadratic formula a = -B ± sqrt(B^2 - 4AC)2A. Here, A=1, B=sqrt(3), C=-3. a = -sqrt(3) ± sqrt((3))^2 - 4(1)(-3)2(1) a = -sqrt(3) ± sqrt(3 + 12)2 a = -sqrt(3) ± sqrt(15)2 Step 5: Choose the valid solution for a. Since a represents a side length, it must be positive. a = -sqrt(3) + sqrt(15)2 The other solution, a = -sqrt(3) - sqrt(15)2, is negative and thus not valid for a side length. Step 6: Manipulate the expression for a to match the required form. a = sqrt(15) - sqrt(3)2 Factor out sqrt(3) from the numerator: a = sqrt(3 × 5) - sqrt(3)2 a = sqrt(3)sqrt(5) - sqrt(3)2 a = sqrt(3)(sqrt(5)-1)2 This matches the required expression. Step 7: Verify the condition a < b < sqrt(3). We have a = sqrt(3)(sqrt(5)-1)2. From b^2 = asqrt(3): b^2 = (sqrt(3)(sqrt(5)-1)2)sqrt(3) b^2 = 3(sqrt(5)-1)2 Since sqrt(5) ≈ 2.236, sqrt(3) ≈ 1.732. a ≈ (1.732(2.236-1))/(2) = (1.732(1.236))/(2) ≈ 1.070 b^2 ≈ (3(2.236-1))/(2) = (3(1.236))/(2) ≈ 1.854 b ≈ sqrt(1.854) ≈ 1.361 The condition a < b < sqrt(3) becomes 1.070 < 1.361 < 1.732, which is true. Thus, we have proven that a = sqrt(3)(sqrt(5)-1)2. Send me the next one 📸