Question

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 the ratio:
$\frac{P_1+P_2+P_3+...}{A_1+A_2+A_3+...}$ equals

• A. $2\frac{1+\sqrt{2}}{a}$
• B. $\frac{2(2-\sqrt{2})}{a}$
• C. $\frac{2(2+\sqrt{2})}{a}$
• D. $\frac{2(1+2\sqrt{2})}{a}$

Answer 1

The correct option is C $\frac{2(2+\sqrt{2})}{a}$

Acc to Ques
Each side of $S_1=a$ units
Each side of $S_2=\sqrt{{\frac{a}{2}}^2+{\frac{a}{2}}^2}=\frac{a}{\sqrt{2}}$ units
Each side of $S_3=\sqrt{\frac{a^2}{2\sqrt{2}}+\frac{a^2}{2\sqrt{2}}}=\frac{a}{2}$ units
Similarly, each side of $S_4=\frac{a}{2\sqrt{2}}$ units
and so on...
Now, $P_1+P_2+P_3+......=4a+4\frac{a}{\sqrt{2}}+\frac{4a}{2}+.....$
It is an infinite G.P series with common ratio $\frac{1}{\sqrt{2}}$
Sum of infinite G.P. series is $\frac{a}{1-r}$, where $a$ is the first term and $r$ is the common ratio
so $P_1+P_2+P_3+......=\frac{4a}{1-\frac{1}{\sqrt{2}}}$
$=\frac{4\sqrt{2}a}{\sqrt{2}-1}$
$=4\sqrt{2}(\sqrt{2}+1)a$ (by rationalising)
Similarly,
$A_1+A_2+A_3+......=a^2+\frac{a^2}{2}+\frac{a^2}{4}+.....$
$=\frac{a^2}{1-\frac{1}{2}}$ (Common Ratio $=\frac{1}{2}$)
$=2a^2$
Now, $\frac{P_1+P_2+P_3+......}{A_1+A_2+A_3+.....}=\frac{4\sqrt{2}(\sqrt{2}+1)a}{2a^2}$
$=\frac{2\sqrt{2}(\sqrt{2}+1)}{a}=\frac{2(\sqrt{2}+2)}{a}$