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Question:

Let S₁be a square of side a. Another square S₂ is formed by joining the mid-points of the sides of S₁. The same process is applied to S₂ to form yet another square S₃, and so on. If A₁, A₂, A₃... are the areas and P₁, P₂, P₃... are the perimeters of S₁, S₂, S₃... respectively,

then the ratio $\frac{P_{1} + P_{2} + P_{3} + . . . . .}{A_{1} + A_{2} + A_{3} + . . .}$ equals:

Explanation:

Area and perimeter of square S₁is a², 4a respectively.

Now, the side of S₂ will be $\frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$

Thus, the area and perimeter of S₂ = $\frac{a^{2}}{2}, \frac{4 a}{\sqrt{2}}$

The side of S₃ will be $\frac{a}{2 \sqrt{2}} \times \sqrt{2} = \frac{a}{2}$

Thus, area and perimeter of S₃ = $\frac{a^{2}}{4}, \frac{4 a}{{(\sqrt{2})}^{2}}$

Similarly, the area and perimeter of S₄ = $\frac{a^{2}}{8}, \frac{4 a}{{(\sqrt{2})}^{3}}$

∴ The required ratio

$\begin{aligned} = \frac{4 a + \frac{4 a}{\sqrt{2}} + \frac{4 a}{(\sqrt{2})^{2}} + \frac{4 a}{(\sqrt{2})^{3}} + \dots}{a^{2} + \frac{a^{2}}{2} + \frac{a^{2}}{4} + \frac{a^{2}}{8} + \dots} \\ = \frac{4 a (1 + \frac{1}{\sqrt{2}} + \frac{1}{(\sqrt{2})^{2}} + \frac{1}{(\sqrt{2})^{3}} \dots)}{a^{2} (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \dots)} \\ = \frac{4 a (\frac{1}{1 - \frac{1}{\sqrt{2}}})}{a^{2} (\frac{1}{1 - \frac{1}{2}})} = \frac{4 a (\frac{\sqrt{2}}{\sqrt{2} - 1})}{a^{2} \times 2} \\ = \frac{4 a \times \sqrt{2} (\sqrt{2} + 1)}{2 a^{2}} = \frac{2 (2 + \sqrt{2})}{a} \end{aligned}$

Hence, option (c).

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