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 $b_{i+1}=\frac{b_i}{2};$ $a_1=0,b_1=32,$ then

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

Answer 1

Answer: (A), (C)

The correct options are
A $A_3=128$
C $\underset{n\to \infty }{lim}\sum _{i=1}^n{A}_i=\frac{8}{3}(32{)}^2$

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 $\sqrt{2}{b}_i.$
$A_i=(\sqrt{2}{b}_i{)}^2=2b_i^2$
$A_{i+1}=2b_{i+1}^2=\frac{2b_i^2}{4}=\frac{A_i}{4}$

$A_1,A_2,A_3,\dots$ form a decreasing G.P. with common ratio $\frac{1}{4}.$
$A_1=2(32{)}^2=2^{11}$
$\therefore A_3=2^{11}\times {\left(\frac{1}{4}\right)}^2=2^7=128$
$\underset{n\to \infty }{lim}\sum _{i=1}^n{A}_i=\frac{2^{11}}{1-\frac{1}{4}}$
$=\frac{2^{13}}{3}=\frac{2^3}{3}(2^5{)}^2$
$=\frac{8}{3}(32{)}^2$