Question

Starting with a unit square, a sequence of square is generated. Each square in the sequence has half the side length of its predecessor and two of its sides bisected by its predecessor's sides as shown. This process is repeated indefinitely. The total area enclosed by all the square in limiting situation, is

• A. $\frac{5}{4}sq.units$
• B. $\frac{79}{64}sq.units$
• C. $\frac{75}{64}sq.units$
• D. $\frac{4}{3}sq.units$

Answer 1

The correct option is D $\frac{4}{3}sq.units$

Let the side of intitial square $=a=1^{st}$ square side.

$2^{nd}$ square side $=\frac{a}{2}$

$3^{rd}$ square side$=\frac{\frac{a}{2}}{2}=\frac{a}{4}$

and so on

area of $1^{st}$ square $=a^2$

area of $2^{nd}$ square $=\frac{a^2}{4}$

Sum $=a^2+\frac{a^2}{4}+\frac{a^2}{16}+.........$ upto infinity

Sum $=a^2[1+\frac{1}{4}+\frac{1}{16}+....]$

Using sum to infinity of $GP$ , we get

Sum $=a^2\times \frac{1}{1-\frac{1}{4}}=\frac{4a^2}{3}$

Since $a=1,$ Sum $=\frac{4}{3}$ sq.units