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This can be found out by using the formula for sum of n terms of an arithmetic progression.

here,

n is not known.

a=2

d=2 (since it's even)

nth term=98 (last even number less than 100)

using formula for nth term,

nth term=a+(n-1)d

98=2+(n-1)2

therefore, n=49

so, sum of n terms=n/2[2*a+(n-1)d]

putting n=49, a=2, and d=2,

sum=2450

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15y ago
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4y ago

2450

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Q: What is the sum of the positive even integers less than one hundred?
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