If `A_1` is the area area bounded by `|x-a_i| + |y|=b_i, i in N` ,where `a_(i+1) =a_i+3/2 b_i and b_(i+1)=b_i/2,a_i=0 and b_i=32`, then

Options

A. `A_(3)=128`
B. `A_(3)=256`
C. `lim_(nto oo) sum_(i=1)^(n) A_(i)=8/3 (32)^(2)`
D. `lim_(nto oo)sum_(i=1)^(n) A_(i)=4/3(16)^(2)`

Answer

A, C

Explanation

`a_(2)=a_(1)+3/2 b_(1)=48, b_(2)=(b_(1))/2=16`
`a_(3) =48+3/2xx16=72, b_(3)=16/2=8`
So, the three loops from `i=1` to `i=3` are alike Now, area of `i^(th)` loop (square)`=1/2 ("diagonal")^(2)`
`A_(i)=1/2(2b_(i))^(2)=2b_(i)^(2)`
`:.(A_(i+1))/(A_(i))=(2(b_(i+1))^(2))/(2(b_(i))^(2))=1/4`
So, the areas from G.P. series
`:.` sum of the G.P. up to infinite term is
`(A_(i))/(1-r)=2(32)^(2)xx1 1/ (1-1/4)=8/3 (32)^(2)` sq. units