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Question:

Let P₁be the circle of radius r. A square Q₁ is inscribed in P₁ such that all the vertices of the square Q₁ lie on the circumference of P₁. Another circle P₂ is inscribed in Q₁. Another Square Q₂ is inscribed in the circle P₂. Circle P₃ is inscribed in the square Q₂ and so on. If S_N is the area between Q_N and P_N+1, where N represents the set of natural numbers, then the ratio of sum of all such S_N to that of the area of the square Q₁ is:

Explanation:

Radius of P₁ = r ⇒A(P₁) = πr²

Diameter = d = 2r

Side of Q₁ = $\frac{2 r}{\sqrt{2}}$ = r $\sqrt{2}$ ⇒ A(Q₁) = 2r²

Radius of P₂ = $\frac{r \sqrt{2}}{2}$ = $\frac{r}{\sqrt{2}}$ ⇒ A(P₁) = $\frac{π r^{2}}{2}$

Side of Q₂ = $(\frac{r}{\sqrt{2}})$ × $\sqrt{2}$ = r ⇒ A(Q₂) = r²... and so on.

i.e., Areas of circles are in G.P. with common ratio = $\frac{1}{2}$

Also, areas of squares are in G.P. with common ratio = $\frac{1}{2}$

S_N = Q₁ – P₂ + Q₂ – P₃ + Q₃ – P₄ + …

= (Q₁ + Q₂ + Q₃ + … ) – (P₂ + P₃ + P₄ + … )

= 2r² $(1 + \frac{1}{2} + \frac{1}{4} + . . .)$ - $\frac{π r^{2}}{2 (1 + \frac{1}{2} + \frac{1}{4} + . . .)}$

= $(1 + \frac{1}{2} + \frac{1}{4} + . . .)$ $(2 r^{2} - \frac{π r^{2}}{2})$

= r² × $[\frac{1}{1 - \frac{1}{2}}]$ × $\frac{4 - π}{2}$

= (4 - π)r²

S_N: A(Q₁) = (4 – π)r²: 2r² = (4 – π)/2

Hence, option (a).

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