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Question:

Let C be a circle with centre P₀ and AB be a diameter of C. Suppose P₁ is the mid-point of the line segment P₀B, P₂ is the mid-point of the line segment P₁B and so on. Let C₁, C₂, C₃, ... be circles with diameters P₀P₁, P₁P₂, P₂P₃ ... respectively. Suppose the circles C₁,C₂, C₃, ... are all shaded. The ratio of the area of the unshaded portion of C to that of the original circle C is:

Explanation:

Let the radius of C be r.

Then P₀P₁ = $\frac{r}{2}$

P₁P₂ = $\frac{r}{4}$

P₂P₃ = $\frac{r}{8}$

and so on.

Thus the areas of circles with diameters P₀P₁, P₁P₂, P₂P₃, … are

$\frac{π r^{2}}{16}, \frac{π r^{2}}{64}, \frac{π r^{2}}{256}, \frac{π r^{2}}{1024} . . . .$

The sum of these areas = $\frac{π r^{2}}{16} \times \frac{1}{1 - \frac{1}{4}} = \frac{π r^{2}}{12}$

∴ Shaded area = $\frac{π r^{2}}{12}$

Unshaded area = πr² - $\frac{π r^{2}}{12} = \frac{11 π r^{2}}{12}$

∴ Required ratio = 11 : 12

Hence, option (d).

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