For an integer , if the arithmetic mean of all coefficient in the binomial expansion of is 16, then the distance of the point from the line x + y = 8 is [2025]

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Solution

Q.
For an integer $n \geq 2$ , if the arithmetic mean of all coefficient 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 [2025]

1 $\sqrt{2}$

2 $2 \sqrt{2}$

3 $3 \sqrt{2}$

4 $5 \sqrt{2}$

Ans.
(3)

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

If x = 1, y = 1

Then, sum of all coefficients = $2^{2 n - 3}$

So, arithmetic mean of all coefficients = $\frac{2^{2 n - 3}}{(2 n - 3) + 1}$

$\Rightarrow \frac{2^{2 n - 3}}{2 n - 2} = 16 \Rightarrow 2^{2 n} = 256 (n - 1) \Rightarrow n = 5$

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

So, distance from P(9, 5) to the line x + y = 8 is

$\frac{(9 + 5 - 8)}{\sqrt{1 + 1}} = \frac{6}{\sqrt{2}} = 3 \sqrt{2} units$ .

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