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Question

The sum of 0.2+0.004+0.00006+0.0000008+ , is

A
200891
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B
20009801
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C
10009801
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D
none of these
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Solution

The correct option is D none of these
Let the sum be S,
S=210+4103+6105+8107(1)
Dividing above equation by 100,
S100=2103+4105+6107+(2)
Subtracting (2) from (1) by shifting one place, we get
99S100=210+2103+2105+2107+
By using formula,
a+ar+ar2+=a1r
Here a=210,r=1100,
We get,
99S100=21011100
S=20009801

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