Let ⁿ Cᵣ₋₁=28, ⁿ Cᵣ=56 and ⁿ Cᵣ₊₁=70. Let A(4 cost, 4 sin t), B(2 sin t,-2 cos t) and C(3 r-n, r²-n-1) be the vertices of a triangle A B C, where t is a parameter. If (3 x-1)²+(3 y)² =α, is the locus of the centroid of triangle ABC , then α equals

Solution

Maths sir · 4 steps

Pin down n and r from the ratios first, then square and add, and the cross terms in t cancel out.

1. Use ratios. Use ⁿCᵣ/ⁿCᵣ₋₁ = (n-r+1)/r and ⁿCᵣ₊₁/ⁿCᵣ = (n-r)/(r+1). These give (n-r+1)/r = 56/28 = 2, so n = 3r - 1. They also give (n-r)/(r+1) = 70/56 = 5/4, so 4n = 9r + 5.
2. Find n and r. Substituting gives 12r - 4 = 9r + 5, so r = 3 and n = 8. Check: ⁸C₂ = 28, ⁸C₃ = 56, ⁸C₄ = 70. So C = (9 - 8, 9 - 8 - 1) = (1, 0).
3. Centroid. With centroid (x, y), we get 3x = 4 cos t + 2 sin t + 1 and 3y = 4 sin t - 2 cos t. So 3x - 1 = 4 cos t + 2 sin t and 3y = 4 sin t - 2 cos t.

   (3x-1)² + (3y)² = (16 + 4)cos² t + (4 + 16)sin² t + (16 - 16)sin t cos t = 20

4. So. The sin t cos t terms cancel and sin² t + cos² t = 1, so the locus is (3x-1)² + (3y)² = 20. That gives α = 20, option (c).

The idea to keep. Three consecutive binomial coefficients give two ratio equations, enough to find n and r. For a locus with a parameter t, isolate the cos t and sin t parts, then square and add to remove t.

Answer from the NTA final answer key. Solution by PrepWiser.