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MCQMediumJEE 2025Basics: Distance, Section Formula, Locus

JEE Mathematics 2025 Question with Solution

Let Tn−1=28T_{n-1} = 28, Tn=56T_n = 56, and Tn+1=70T_{n+1} = 70. Let A (4cos⁡t,4sin⁡t)\left(4\cos t, 4\sin t\right), B (2sin⁡t,−2cos⁡t)\left(2\sin t, -2\cos t\right), and C (3rn−1,rn2−n−1)\left(3r_n - 1, r_n^2 - n - 1\right) be the vertices of a triangle ABC, where tt is a parameter. If (3x−1)2+(3y)2=a\left(3x - 1\right)^2 + \left(3y\right)^2 = a, is the locus of the centroid of triangle ABC, then aa equals:

  • A

    1818

  • B

    88

  • C

    66

  • D

    2020

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Tn−1=28T_{n-1} = 28, Tn=56T_n = 56, Tn+1=70T_{n+1} = 70. The vertices are A(4cos⁡t,4sin⁡t)A\left(4\cos t, 4\sin t\right), B(2sin⁡t,−2cos⁡t)B\left(2\sin t, -2\cos t\right) and, from the solution working, C(3r−n,r2−n−1)C\left(3r-n, r^2-n-1\right).

Find: The value of aa in the locus (3x−1)2+(3y)2=a\left(3x-1\right)^2 + \left(3y\right)^2 = a for the centroid of triangle ABCABC.

From the binomial coefficient relations shown in the solution:

nCr−1nCr=2856=12\frac{{}^nC_{r-1}}{{}^nC_r} = \frac{28}{56} = \frac{1}{2}

which gives

rn−r+1=12\frac{r}{n-r+1} = \frac{1}{2}

and hence

3r=n+1...(i)3r = n+1 \qquad \text{...(i)}

Using the next pair,

nCrnCr+1=5670=45\frac{{}^nC_r}{{}^nC_{r+1}} = \frac{56}{70} = \frac{4}{5}

we get

r+1n−r=45\frac{r+1}{n-r} = \frac{4}{5}

Together with 3r=n+13r=n+1, the solution gives

r=3,n=8r=3, \quad n=8

Coordinate Simplification

Substituting r=3r=3 and n=8n=8 in point CC:

C=(3r−n,r2−n−1)=(9−8,9−8−1)=(1,0)C = \left(3r-n, r^2-n-1\right) = \left(9-8, 9-8-1\right) = \left(1,0\right)

So the triangle vertices become

A(4cos⁡t,4sin⁡t),B(2sin⁡t,−2cos⁡t),C(1,0)A\left(4\cos t, 4\sin t\right), \quad B\left(2\sin t, -2\cos t\right), \quad C\left(1,0\right)

If the centroid is (x,y)\left(x,y\right), then

3x=4cos⁡t+2sin⁡t+13x = 4\cos t + 2\sin t + 1 3y=4sin⁡t−2cos⁡t3y = 4\sin t - 2\cos t

Therefore,

3x−1=4cos⁡t+2sin⁡t3x - 1 = 4\cos t + 2\sin t 3y=4sin⁡t−2cos⁡t3y = 4\sin t - 2\cos t

Now,

(3x−1)2+(3y)2=(4cos⁡t+2sin⁡t)2+(4sin⁡t−2cos⁡t)2=16cos⁡2t+16sin⁡tcos⁡t+4sin⁡2t+16sin⁡2t−16sin⁡tcos⁡t+4cos⁡2t=20(sin⁡2t+cos⁡2t)=20\begin{aligned} \left(3x-1\right)^2 + \left(3y\right)^2 &= \left(4\cos t + 2\sin t\right)^2 + \left(4\sin t - 2\cos t\right)^2 \\ &= 16\cos^2 t + 16\sin t \cos t + 4\sin^2 t \\ &\quad + 16\sin^2 t - 16\sin t \cos t + 4\cos^2 t \\ &= 20\left(\sin^2 t + \cos^2 t\right) \\ &= 20 \end{aligned}

Hence a=20a=20. Therefore, the correct option is D. The solution marks option C, but its own working leads to a=20a=20, so the working is taken as authoritative.

Common mistakes

  • Using the listed correct answer without checking the algebra. The solution working gives (3x−1)2+(3y)2=20\left(3x-1\right)^2 + \left(3y\right)^2 = 20, so the answer must follow the derivation, not the mislabeled option.

  • Computing the centroid incorrectly by forgetting to divide each coordinate sum by 33. For a triangle, centroid coordinates are the averages of the three vertex coordinates.

  • Substituting point CC incorrectly. After solving, r=3r=3 and n=8n=8, so C=(1,0)C=(1,0). Using the question text expression (3rn−1,rn2−n−1)\left(3r_n-1, r_n^2-n-1\right) directly leads to confusion; the solution uses (3r−n,r2−n−1)\left(3r-n, r^2-n-1\right).

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