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Question

Find the sum of all the odd positive integers less than 100.

A
2400
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B
2500
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C
2525
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D
2600
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E
2650
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Solution

The correct option is B 2500
The possible odd positive integers: 1,3,5,7,9,11..........99
The first term, a =1
The difference between two consecutive terms =31=2
The total no. of terms =50
Applying sum of arithmetic progression formula,
Sn=n2[2a+(n1)d]
=502[2×1+(501)2]
=25(2+98)=2500
Hence, choice B is correct.

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