Question

If ‘$A_i$’ is the area bounded by bounded by $|x-a_i|+|y|=b_i,iϵN,where{a}_{i+1}=a_i+\frac{3}{2}{b}_{i}and{b}_{i+1}=\frac{b_i}{2},a_1=0,b_1=32$ then

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

Answer 1

Answer: (A), (C)

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

$a_1=0,b_1=32,a_1+\frac{3}{2}{b}_1=48,\frac{b_1}{2}=16,a_3=48+\frac{3}{2}\times 16=72,b_3=\frac{16}{2}=8$

So the three loops from i = 1 to i = 3 are alike
Now area of $i^{th}$ loop (square) $=\frac{1}{2}(diagonal{)}^2$
$A_i=\frac{1}{2}(2b_i{)}^2=2(b_i{)}^2$
$So,\frac{A_{i+1}}{2(b_i{)}^2}=\frac{1}{4}$
So, the sum of the G.P. upto infinite terms $=A_i\left(\frac{1}{1-r}\right)=2(32{)}^2\times \frac{1}{1-\frac{1}{4}}$
$=2(32{)}^2\times \left(\frac{4}{3}\right)=\frac{8}{3}\times (32{)}^2$ square units.