Question

$P_1,P_2,\dots \dots .,P_n$ are points on the line $y=x$ lying in the positive quadrant such that $O{P}_n=n\cdot O{P}_{n-1}$, where $O$ is the origin. If $O{P}_1=1$ and the coordinates of $P_n$ are $(2520\sqrt{2},2520\sqrt{2})$, then $n$ is equal to

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is C $7$

$P_1,P_2.P_3....$ are points on line $y=x$

$O{P}_1=1$

Let $P_1=(x,x$)

But $O{P}_1=1\Rightarrow 2x^2=1\Rightarrow x^2=\frac{1}{2}\Rightarrow x=\frac{1}{\sqrt{2}}$

So $P_1=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

$O{P}_2=2\times O{P}_1=2\Rightarrow 2x^2=4\Rightarrow x=\sqrt{2}\Rightarrow P_2=(\sqrt{2},\sqrt{2})$

$O{P}_3=2\times 3=6\Rightarrow 2x^2=6^2\Rightarrow x=3\sqrt{2}\Rightarrow P_3=(3\sqrt{2},3\sqrt{2})$

....coninuing this way get

$O{P}_7=7\times 720=5040\Rightarrow 2x^2=5040\times 5040\Rightarrow x^2=\Rightarrow x=2520\sqrt{2}\Rightarrow P_7=(2520\sqrt{2},2520\sqrt{2})$

Comparing with given point we get $n=7$.