A 1 , A 2 , … , A n are points on the line y = x lying in the positive quadrant such that O A n = n · O A n - 1 , O being the origin. If O A 1 = 1 and the coordinates of A n are ( 2520 2 , ? 2520 2 ) , then the value of n is

Question

$A_{1}, A_{2}, \dots, A_{n}$ are points on the line $y = x$ lying in the positive quadrant such that $O A_{n} = n \cdot O A_{n - 1},$ $O$ being the origin. If $O A_{1} = 1$ and the coordinates of $A_{n}$ are $(2520 \sqrt{2}, 2520 \sqrt{2})$ , then the value of $n$ is _______.

Smart Achievers · Accepted Answer

(7)

$We have, O A_{n} = n \cdot O A_{n - 1}$

$= n (n - 1) \cdot O A_{n - 2} = \dots = n! \cdot O A_{1} = n!$

$∴$ $O A_{n} = n!$

$\Rightarrow \sqrt{2} \times 2520 \sqrt{2} = n!$

$\Rightarrow n! = 5040$

$\Rightarrow n = 7$

Solution

Q.
$A_{1}, A_{2}, \dots, A_{n}$ are points on the line $y = x$ lying in the positive quadrant such that $O A_{n} = n \cdot O A_{n - 1},$ $O$ being the origin. If $O A_{1} = 1$ and the coordinates of $A_{n}$ are $(2520 \sqrt{2}, 2520 \sqrt{2})$ , then the value of $n$ is _______.

Ans.
(7)

$We have, O A_{n} = n \cdot O A_{n - 1}$

$= n (n - 1) \cdot O A_{n - 2} = \dots = n! \cdot O A_{1} = n!$

$∴$ $O A_{n} = n!$

$\Rightarrow \sqrt{2} \times 2520 \sqrt{2} = n!$

$\Rightarrow n! = 5040$

$\Rightarrow n = 7$

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Solution

Q. A1,A2,…,An are points on the line y=x lying in the positive quadrant such that OAn=n·OAn-1, O being the origin. If OA1=1 and the coordinates of An are (25202, 25202), then the value of n is _______.

Ans. (7) We have, OAn=n·OAn-1 =n(n-1)·OAn-2=…=n!·OA1=n! ∴ OAn=n! ⇒2×25202=n! ⇒n!=5040 ⇒n=7

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Q.
$A_{1}, A_{2}, \dots, A_{n}$ are points on the line $y = x$ lying in the positive quadrant such that $O A_{n} = n \cdot O A_{n - 1},$ $O$ being the origin. If $O A_{1} = 1$ and the coordinates of $A_{n}$ are $(2520 \sqrt{2}, 2520 \sqrt{2})$ , then the value of $n$ is _______.

Ans.
(7)

$We have, O A_{n} = n \cdot O A_{n - 1}$

$= n (n - 1) \cdot O A_{n - 2} = \dots = n! \cdot O A_{1} = n!$

$∴$ $O A_{n} = n!$

$\Rightarrow \sqrt{2} \times 2520 \sqrt{2} = n!$

$\Rightarrow n! = 5040$

$\Rightarrow n = 7$