Question

If $A_i$ is the area bounded by $|x-a_i|+|y|=b_i,i\in N,$ 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)

$\underset{n\to \infty }{lim}\sum _{i=1}^n{A}_i=\frac{8}{3}(32{)}^2$


$a_1=0,b_1=32,$
$a_2=a_1+\frac{3}{2}{b}_1=48,b_2=\frac{b_1}{2}=16$
$a_3=48+\frac{3}{2}\times 16=72,b_3=\frac{16}{2}=8$
So, the three loops from $i=1$ to $i=3$ are alike.
Now, area of $i\text{th}$ loop (square) $=\frac{1}{2}(\text{diagonal}{)}^2$
$A_i=\frac{1}{2}(2b_i{)}^2=2(b_i{)}^2$

So, $\frac{A_{i+1}}{A_i}=\frac{2(b_{i+1}{)}^2}{2(b_i{)}^2}=\frac{1}{4}$

So, the areas form a G.P. series.
So, the sum of the G.P. up to infinite terms is
$A_1\frac{1}{1-r}=2(32{)}^2\times \frac{1}{1-\frac{1}{4}}$
$=2\times (32{)}^2\times \frac{4}{3}$
$=\frac{8}{3}(32{)}^2\text{sq. units}$