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If A(i) is the area bounded by |x-a(i)|+...

If `A_(i)` is the 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_(1)=0,=b_(1)=32,` then

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)`

Text Solution

Verified by Experts

The correct Answer is:
A, C
`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
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