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$A_{1}, A_{2}, A_{3} ....... A_{n}$ 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 $O A_{1} = 1$ then the coordinates of $A_{8}$ are $(3 a \sqrt{2}, 3 a \sqrt{2})$ then $a =$

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answer is 6720.

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Detailed Solution

$O A_{n} = n O . A_{n - 1}$

$O A_{2} = 2 O A_{1} = 2$

$O A_{3} = 3 O A_{2} = 3 \times 2 = 6$

$O A_{4} = 4. O A_{3} = 4 \times 6 = 24$

$O A_{5} = 120, O A_{6} = 720$

$O A_{7} = 5080, O A_{8} = 40320$

$O A_{8} = \sqrt{2} x = 40320 \Rightarrow x = \frac{20160 \times 2}{\sqrt{2}} = \sqrt{2} (20160)$

$\begin{array}{l} A_{8} = (20160 \sqrt{2}, 20160 \sqrt{2}) \\ A_{8} (3 a \sqrt{2}, 3 a \sqrt{2}) \end{array}$

$\begin{array}{l} 3 a \sqrt{2} = 20160 \sqrt{2} \\ a = 6720 \end{array}$

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