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Question:

Let A₁ be a square whose side is ‘a’ metres. Circle C₁ circumscribes the square A₁ such that all its vertices are on C₁. Another square A₂ circumscribes C₁. Circle C₂ circumscribes A₂, and A₃ circumscribes C₂, and so on. If D_N is the area between the square A_N and the circle C_N, where N is a natural number, then the ratio of the sum of all D_N to D₁ is:

Explanation:

Area of square 1 = a²

Area of circle 1 = π ${(\frac{a}{\sqrt{2}})}^{2}$

∴ D₁ = $\frac{a^{2}}{2}$ (π - 2)

Area of square 2 = (a $\sqrt{2}$ )²

Area of circle 2 = πa²

∴ D₂ = a²(π - 2)

Similarly,

D₃ = 2a² (π - 2), and so on.

∴ D₁ + D₂ + D₃ + ... = a²(π - 2) [0.5 + 1 + 2 + 4 + ...]

= Infinite

Hence, option (c).

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