Let $S_{1}$ be a square of side a. Another square $S_{2}$ is formed by joining the mid-points of the sides of $S_{1}$ . The same process is applied to $S_{2}$ to form yet another square $S_{3}$ , and so on. If $A_{1}, A_{2}, A_{3} ........$ be the areas and $P_{1}, P_{2}, P_{3} .........$ be the perimeters of $S_{1}, S_{2}, S_{3} .......$ , respectively then, compute $\lim_{n \to \infty} \frac{P_{1} + P_{2} + P_{3} + .... + P_{n}}{A_{1} + A_{2} + A_{3} + .... + A_{n}}$ .

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$\frac{2}{a} (2 + \sqrt{2})$

$\frac{2}{a} (2 - \sqrt{2})$

$2 a (2 + \sqrt{2})$

$2 a (2 - \sqrt{2})$

answer is A.

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Detailed Solution

$A_{1} = a^{2}, P_{1} = 4 a$
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$A_{2} = \frac{a^{2}}{2}, P_{2} = \frac{4 a}{\sqrt{2}}, A_{3} = \frac{a^{2}}{4}, P_{3} = \frac{4 a}{{(\sqrt{2})}^{2}}$ and so on
$∴ \lim_{n \to \infty} \frac{P_{1} + P_{2} + P_{3} + .. + P_{n}}{A_{1} + A_{2} + A_{3} + ... + A_{n}} = \frac{4 a + \frac{4 a}{\sqrt{2}} + \frac{4 a}{{(\sqrt{2})}^{2}} + \frac{4 a}{{(\sqrt{2})}^{3}} + ... \infty}{a^{2} + \frac{a^{2}}{2} + \frac{a^{2}}{4} + \frac{a^{2}}{8} + .... \infty} = \frac{4 a (\frac{1}{1 - \frac{1}{\sqrt{2}}})}{a^{2} (\frac{1}{1 - \frac{1}{2}})}$
(Sum of infinite $G . P . = \frac{a}{1 - r}$ where a is first term of G.P. and r is common ration.)

$= \frac{2}{a} (\frac{\sqrt{2}}{\sqrt{2} - 1}) = \frac{2}{a} (2 + \sqrt{2})$