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Mathematics · Binomial Theorem

JEE Main 2025 — 4 April, Morning Shift — Question 34

For an integer n≥2n \geq 2, if the arithmetic mean of all coefficients in the binomial expansion of (x+y)2n−3(x+y)^{2 n-3} is 16 , then the distance of the point P(2n−1P(2 n-1, n2−4nn^{2}-4 n ) from the line x+y=8x+y=8 is

  1. Option A:

    222 \sqrt{2}

  2. Option B:

    2\sqrt{2}

  3. Option C:

    525 \sqrt{2}

  4. Option D:

    323 \sqrt{2}

    Correct

Answer: D

Step-by-step solution

 Mean =2n−3C0+2n−3C1+2n−3C2+⋯2n−3C2n−32n−2=16=22n−3=16(2n−2)=22n−3=25(n−1)⇒n=5∴P(9,5)d=∣9+5−82∣=62=32\begin{aligned} \text { Mean } & =\frac{{ }^{2 n-3} C_{0}+{ }^{2 n-3} C_{1}+{ }^{2 n-3} C_{2}+\cdots{ }^{2 n-3} C_{2 n-3}}{2 n-2}=16 \\& =2^{2 n-3}=16(2 n-2) \\& =2^{2 n-3}=2^{5}(n-1) \\& \Rightarrow n=5 \\& \therefore \quad P(9,5) d \\& =\left|\frac{9+5-8}{\sqrt{2}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem
Topic
Binomial Coefficients

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