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A square is drawn by joining mid points of the sides of a square. Another square is drawn inside the second square in the same way and the process is continued in definitely. If the side of the first square is $16 \mathrm{~cm}$, then what is the sum of the areas of all the squares?
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The correct answer is:
$512 \mathrm{sq} \mathrm{cm}$
Let ABCD, EFGH and IJKL be squares. Let side of square $\mathrm{ABCD}=16$ Now, Area of $\mathrm{ABCD}=(16)^{2}$
Area of EFGH $=\frac{(16)^{2}}{2}$,
Area of IJKL $=\frac{(16)^{2}}{4}$ So on.
Required sum,
$=16^{2}+\frac{1}{2}(16)^{2}+\frac{1}{4}(16)^{2}+\ldots . \infty$
$=(16)^{2}\left\{1+\frac{1}{2}+\frac{1}{4}+\ldots \infty\right\}$
Now, $1+\frac{1}{2}+\frac{1}{4}+\ldots \ldots \infty$ is a G.P.
$\therefore \quad$ Sum $=\frac{a}{1-r}$ where $a=1$ and $r=\frac{1}{2}$
$\therefore \quad 1+\frac{1}{2}+\frac{1}{4}+\ldots \ldots+\infty=\frac{1}{1-\frac{1}{2}}=2$
$=256 \times 2=512$ sq. $\mathrm{cm}$
Area of EFGH $=\frac{(16)^{2}}{2}$,
Area of IJKL $=\frac{(16)^{2}}{4}$ So on.
Required sum,
$=16^{2}+\frac{1}{2}(16)^{2}+\frac{1}{4}(16)^{2}+\ldots . \infty$
$=(16)^{2}\left\{1+\frac{1}{2}+\frac{1}{4}+\ldots \infty\right\}$
Now, $1+\frac{1}{2}+\frac{1}{4}+\ldots \ldots \infty$ is a G.P.
$\therefore \quad$ Sum $=\frac{a}{1-r}$ where $a=1$ and $r=\frac{1}{2}$
$\therefore \quad 1+\frac{1}{2}+\frac{1}{4}+\ldots \ldots+\infty=\frac{1}{1-\frac{1}{2}}=2$
$=256 \times 2=512$ sq. $\mathrm{cm}$
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