Sequences & Series
Geometric Progression
Grade 11
Question:
<p>\(A_r, r = 1, 2, 3, \ldots, n\), are \(n\) points on the parabola \(y^2 = 4x\) in the first quadrant. If \(A_r = (x_r, y_r)\), where \(x_1, x_2, x_3, \ldots, x_n\) are in GP and \(x_1 = 1\), \(x_2 = 2\), then \(y_n\) is equal to</p>
<p>\(y_{11} = 2^6\)</p>
<p>\(y_{17} = 2^9\)</p>
<p>\(y_{22} = 2^{\frac{23}{2}}\)</p>
<p>\(y_{18} = 2^9\)</p>
Step-by-Step Solution
Key Concept: Since points lie on parabola y² = 4x, once we find the common ratio of the GP for x-coordinates, we can determine xₙ and then use y²ₙ = 4xₙ to find yₙ. The constraint that points are in the first quadrant means yₙ > 0.
<p><strong>Step 1:</strong> Identify the common ratio of GP. Given x₁ = 1 and x₂ = 2, the common ratio is r = x₂/x₁ = 2/1 = 2.</p><p><strong>Step 2:</strong> Find xₙ using GP formula: xₙ = x₁ · r^(n-1) = 1 · 2^(n-1) = 2^(n-1).</p><p><strong>Step 3:</strong> Since point Aₙ = (xₙ, yₙ) lies on parabola y² = 4x, substitute: y²ₙ = 4xₙ = 4 · 2^(n-1) = 2² · 2^(n-1) = 2^(n+1).</p><p><strong>Step 4:</strong> Since the point is in the first quadrant, yₙ > 0. Therefore: yₙ = √(2^(n+1)) = 2^((n+1)/2).</p><p>∴ Answer: B</p>
Correct Answer: B