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Byju's Answer

Standard XII

Mathematics

Permutation

A1,A2,....An ...

Question

$A_{1}, A_{2}, . . . . A_{n}$ are point on the line $y = x$ lying in the positive quadrant such that $O A_{n} = n O A_{n - 1}$ , $O$ is origin. If $O A_{1} = 1$ and $A_{n} (2520 \sqrt{2}, 2520 \sqrt{2})$ then $n =$

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Solution

The correct option is B $7$
$A_{n} (2520 \sqrt{2}, 2520 \sqrt{2})$
$O A_{n} = 2520 \sqrt{2} \times \sqrt{2} = 5040$
${O A}_{1} = 1$
${O A}_{2} = 2.1 = 2!$
${O A}_{3} = 3.2.1 = 3!$
$.$
$.$
$.$
${O A}_{n} = n!$
$\Rightarrow n! = 5040 = 7!$
$∴$ $n = 7$ [C]

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A1,A2,....An are point on the line y=x lying in the positive quadrant such that OAn=nOAn−1, O is origin. If OA1=1 and An(2520√2,2520√2) then n=

The correct option is B 7An(2520√2,2520√2)OAn=2520√2×√2=5040OA1=1OA2=2.1=2!OA3=3.2.1=3!...OAn=n!⇒n!=5040=7!∴ n=7 [C]

$A_{1}, A_{2}, . . . . A_{n}$ are point on the line $y = x$ lying in the positive quadrant such that $O A_{n} = n O A_{n - 1}$ , $O$ is origin. If $O A_{1} = 1$ and $A_{n} (2520 \sqrt{2}, 2520 \sqrt{2})$ then $n =$

The correct option is B $7$
$A_{n} (2520 \sqrt{2}, 2520 \sqrt{2})$
$O A_{n} = 2520 \sqrt{2} \times \sqrt{2} = 5040$
${O A}_{1} = 1$
${O A}_{2} = 2.1 = 2!$
${O A}_{3} = 3.2.1 = 3!$
$.$
$.$
$.$
${O A}_{n} = n!$
$\Rightarrow n! = 5040 = 7!$
$∴$ $n = 7$ [C]