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Question

If Ai is the area bounded by |xai|+|y|=bi, iN, where ai+1=ai+32bi and bi+1=bi2, a1=0,b1=32, then

A
A3=128
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B
A3=256
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C
limnni=1Ai=83(32)2
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D
limnni=1Ai=43(16)2
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Solution

The correct option is C limnni=1Ai=83(32)2

a1=0, b1=32,
a2=a1+32b1=48, b2=b12=16
a3=48+32×16=72, b3=162=8
So, the three loops from i=1 to i=3 are alike.
Now, area of ith loop (square) =12(diagonal)2
Ai=12(2bi)2=2(bi)2

So, Ai+1Ai=2(bi+1)22(bi)2=14

So, the areas form a G.P. series.
So, the sum of the G.P. up to infinite terms is
A111r=2(32)2×1114
=2×(32)2×43
=83(32)2 sq. units

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