JEE Mathematics 2025 Question with Solution

Given: The arithmetic mean of all coefficients in the expansion of $(x+y)^{2n-3}$ is $16$.

Find: The distance of the point $P(2n-1, n^2-4n)$ from the line $x+y=8$.

For $(x+y)^m$, the sum of coefficients is

$$2^m$$

and the number of coefficients is

$$m+1$$

Hence the arithmetic mean of the coefficients is

$$\frac{2^m}{m+1}$$

Here,

$$m=2n-3$$

So,

$$\frac{2^{2n-3}}{(2n-3)+1}=16$$

that is,

$$\frac{2^{2n-3}}{2n-2}=16=2^4$$

Therefore,

$$2^{2n-7}=2n-2$$

Checking the integer value $n \ge 2$, we get

$$n=5$$

Now the point becomes

$$P(2n-1, n^2-4n)=P(2\cdot 5-1, 5^2-4\cdot 5)=P(9,5)$$

The line is

$$x+y-8=0$$

Using the distance formula from point $(x_0,y_0)$ to line $ax+by+c=0$,

$$\text{Distance}=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$

So,

$$\text{Distance}=\frac{|1\cdot 9+1\cdot 5-8|}{\sqrt{1^2+1^2}}$$ $$=\frac{|14-8|}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}$$

Therefore, the required distance is $3\sqrt{2}$. The correct option is D.

Given: The arithmetic mean of all coefficients in $(x+y)^{2n-3}$ is $16$.

Find: The distance of $P(2n-1, n^2-4n)$ from $x+y=8$.

1. Number of terms in $(x+y)^{2n-3}$ is

$$(2n-3)+1=2n-2$$

2. Sum of all coefficients is

$$2^{2n-3}$$

3. Therefore, arithmetic mean is

$$\frac{2^{2n-3}}{2n-2}=16$$

4. Rewrite as

$$2^{2n-3}=16(2n-2)$$

Since $16=2^4$,

$$2^{2n-3}=2^4(n-1)\cdot 2$$

which is equivalent to

$$2^{2n-7}=2n-2$$

Now checking integer values, $n=5$ satisfies the equation.

5. Substitute $n=5$ in the point:

$$2n-1=9, \qquad n^2-4n=25-20=5$$

So the point is

$$P(9,5)$$

6. Distance from $P(9,5)$ to $x+y=8$ is distance from $P(9,5)$ to

$$x+y-8=0$$

Thus,

$$\text{Distance}=\frac{|9+5-8|}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}$$

Therefore, the correct option is D.