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8. Sequences and Series
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Let $S_1$ be the sum of areas of the squares whose sides are parallel to coordinate axes. Let $S_2$ be the sum of areas of the slanted squares as shown in the figure. Then, $\frac{S_1}{S_2}$ is equal to
A
$2$
B
$\sqrt{2}$
C
$1$
D
$\frac{1}{\sqrt{2}}$
(KVPY-2016)
Solution
(a)
Here,
$S_1=a^2+\left(\frac{a}{2}\right)^2+\left(\frac{a}{4}\right)^2+\ldots$
$S_2=\left(\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{2 \sqrt{2}}\right)^2+\ldots$
$S_1=a^2+\frac{a^2}{4}+\frac{a^2}{16}+\ldots=\frac{a^2}{1-\frac{1}{4}}=\frac{4 a^2}{3}$
$S_2=\frac{a^2}{2}+\frac{a^2}{8}+\frac{a^2}{32}+\ldots=\frac{a^2 / 2}{1-\frac{1}{4}}=\frac{4 a^2}{6}$
$\therefore \quad \frac{S_1}{S_2}=\frac{\frac{4 a^2}{3}}{4 a^2 / 6}=2$
Standard 11
Mathematics
