Question

Let \(T_{n-1} = 28\), \(T_n = 56\), and \(T_{n+1} = 70\). Let A \((4\cos t, 4\sin t)\), B \((2\sin t, -2\cos t)\), and C \((3r_n - 1, r^2_n - n - 1)\) be the vertices of a triangle ABC, where \(t\) is a parameter. If \((3x - 1)^2 + (3y)^2 = a\), is the locus of the centroid of triangle ABC, then \(a\) equals:

• A. 18
• B. 8
• C. 6
• D. 20

Hint:

Review properties of centroids and binomial coefficients to simplify the problem-solving process.

Answer 1

The Correct Option is C

Step 1: Solve for \(n\) from given \(T\) values. Using the property of binomial coefficients: \(T_n = \binom{n}{r}\). Solving \( \binom{n-1}{r} = 28, \binom{n}{r} = 56, \binom{n+1}{r} = 70 \) gives \(n = 8\) and \(r = 3\). 
Step 2: Find centroid coordinates. Centroid \(G\) of triangle ABC has coordinates: \[ G = \left(\frac{4\cos t + 2\sin t + 3r_n - 1}{3}, \frac{4\sin t - 2\cos t + r^2_n - n - 1}{3}\right) \] 
Step 3: Express \(x\) and \(y\) in terms of \(t\) and simplify. Insert values and simplify the coordinates to express \(x\) and \(y\) as functions of \(t\). 
Step 4: Derive the equation of the locus. Substitute \(x\) and \(y\) into \((3x - 1)^2 + (3y)^2 = a\) and simplify to find \(a\). 
Conclusion: After solving, \(a = 6\).