JEE Main 2025 Apr 4 Shift 1, Mathematics Q4: Properties Of Binomial Coefficient

JEE Main2025Apr 4, Shift 1Mathematics

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 $P (2 n - 1, n^{2} - 4 n)$ from the line $x + y = 8$ is:

A
$2$
B
$22$
C
$52$
D
$32$

Solution

$(x + y)^{2 n - 3}$ has $(2 n - 2)$ terms; setting $x = y = 1$ gives the sum of all coefficients as $2^{2 n - 3}$ .

Arithmetic mean of coefficients $= \frac{2 ^{2 n - 3}}{2 n - 2} = 16$ , i.e. $2^{2 n - 3} = 16 (2 n - 2) = 2^{5} (n - 1)$ .

Trying $n = 5$ : LHS $= 2^{7} = 128$ , RHS $= 32 \times 4 = 128$ . So $n = 5$ .

Then $P = (2 n - 1, n^{2} - 4 n) = (9, 5)$ .

Distance from $P$ to $x + y - 8 = 0$ is $\frac{∣9 + 5 - 8∣}{2} = \frac{6}{2} = 32$ .

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JEE Main 2025 Apr 4 Shift 1, Mathematics Q4: Properties Of Binomial Coefficient

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