Question

If $A_i$ is the area bounded by $|x-a_i|+|y|=b_i$, where $a_{i+1}=a_i+\frac{3}{2}{b}_i$ and $b_{i+1}=\frac{b_i}{2},b_1=32,$ then

• A. $\underset{n\to \infty }{lim}\sum _{i=1}^n{A}_i=\frac{4}{3}(16{)}^2$
• B. $\underset{n\to \infty }{lim}\sum _{i=1}^n{A}_i=\frac{8}{3}(32{)}^2$
• C. $A_3=256$
• D. $A_3=128$

Answer 1

Answer: (B), (D)

$A_3=128$

Let the curve $|x-a|+|y-b|=c$ represents the square centred at $(a,b)$ with side length $\sqrt{2}c$
So, $|x-a_i|+|y|=b_i$ is a square centred at $(a_i,0)$ with side length $\sqrt{2}{b}_i$
And if we change the centre of the above square, then bounded area remains same.
$A_i=(\sqrt{2}{b}_i{)}^2=2b_i^2$
$\Rightarrow A_{i+1}=2b_{i+1}^2=\frac{2b_i^2}{4}=\frac{A_i}{4}$
So, $A_1,A_2,A_3,\dots$ form a decreasing $G.P.$ with commom ratio $\frac{1}{4}$
$\Rightarrow A_1=2(32{)}^2=2^{11}$
$\therefore A_3=2^{11}\times {\left(\frac{1}{4}\right)}^2=2^7=128$

Now, $\underset{n\to \infty }{lim}\sum _{i=1}^n{A}_i=\frac{2^{11}}{1-\frac{1}{4}}$
$\Rightarrow \frac{2^{13}}{3}=\frac{8}{3}(2^5{)}^2$
$=\frac{8}{3}(32{)}^2$