The length of a side of a square is ‘a’ meter. A second square is formed by joining the middle points of this square. Then a third square is formed by joining the middle points of the second square and so on. Then the sum of the area of the squire which carried up to infinity is

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a^{2}

2 a^{2}

3 a^{2}

4 a^{2}

answer is B.

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Detailed Solution

Given side of square =

a

The area of square side a =

a^{2}

The side of the square formed by joining the midpoints of original square=

\frac{a}{\sqrt{2}}

Similarly, the second area square =

\frac{a^{2}}{2}

Also, the area of third square which is formed in the same way =

\frac{a^{2}}{2}

Here, we get,
Sides are

a,

\frac{a}{\sqrt{2}}, \frac{a}{2}, \frac{a}{2 \sqrt{2}}

Also, the area of these sides,

a^{2}, \frac{a^{2}}{2}, \frac{a^{2}}{4}, \frac{a^{2}}{8}

Here, the sum of areas of each of those squares is an infinite geometric series in which
First term a= a2 Common ratio r= 1/2
So, the sum of this series,

S_{\infty} = \frac{a}{1 - r}

= \frac{a^{2}}{(1 - \frac{1}{2})}

= 2 a^{2}

So, the required sum is

2 a^{2} .

Hence, option 2 is correct.

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