JEE MAIN - Mathematics (2025 - 4th April Morning Shift - No. 14)
For an integer $n \geq 2$, if the arithmetic mean of all coefficients in the binomial expansion of $(x+y)^{2 n-3}$ is 16 , then the distance of the point $\mathrm{P}\left(2 n-1, n^2-4 n\right)$ from the line $x+y=8$ is
$\sqrt{2}$
$2 \sqrt{2}$
$5 \sqrt{2}$
$3 \sqrt{2}$
Explanation
$$\begin{aligned}
& \text { Mean }=\frac{{ }^{2 n-3} C_0+{ }^{2 n-3} C_1+{ }^{2 n-3} C_2+\cdots{ }^{2 n-3} C_{2 n-3}}{2 n-2}=16 \\
&=2^{2 n-3}=16(2 n-2) \\
&=2^{2 n-3}=2^5(n-1) \\
& \Rightarrow n=5 \\
& \therefore \quad P(9,5) \\
& d=\left|\frac{9+5-8}{\sqrt{2}}\right|=\frac{6}{\sqrt{2}}=3 \sqrt{2}
\end{aligned}$$
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