What is the difference between the sum of the even integers from 1–100

Question

What is the difference between the sum of the even integers from 1–100 and the sum of the odd integers from 1–100?

A 0
B 25
C 26
D 50
E 52

philipssonicare · Answer

They are in pairs
1 with 2
3 with 4
...
99 with 100

Each pair has a difference of 1. There are 50 such pairings.  D

LevanKhukhunashvili · Answer

First even integer 2, last one 100
 \frac{100-2}{2} +1=50 even integers
Average of these 50 even integers  \frac{100+2}{2} =51
Sum of even integers=average*number of even int=51*50

First odd integer 1, last one 99
 \frac{99-1}{2} +1=50 odd integers
average of these odd integers  \frac{99+1}{2} =50
Sum of odd integers=average*number of odd int=50*50

difference between the sum of even and odd integers
51*50-50*50=50(51-50)=50

Imo
Ans:  D

ScottTargetTestPrep · Answer

If we compare the first few even and odd integers starting from 1, we have:

1 and 2

3 and 4

5 and 6

7 and 8

Notice that each even integer is “one more” than its odd counterpart. Since there are 50 even integers and 50 odd integers between 1 and 100, the sum of the even integers will be 50 more than the sum of the odd integers.

Answer: D

KanishkM · Answer

1 to 100, 50 even & 50 odd

Apart from that 1st even number 2 to 100

1st odd number 1 to last odd number 99

Sn = n/2 (1st term + last term), since the series will be an AP

Sn(even) = 50/2 * 102 
= 50 * 51

Sn(odd) = 50/2 * 100
= 50 * 50

Difference = 50 * (51 - 50)
=50

D

Ish1996 · Answer

I dont understand why this doesnt work:

sum of first n even numbers = n(n+1)

sum of first odd numbers = n^2

100(101) - 100^2 = 100

Can someone explain why this formula wont work?

sahibr · Answer

The reason why this method is not working is because the value of n that you have assumed is incorrect.

The even integers form the following series: 2,4,6,8...100
Hence it is an AP with common difference 2
We know in an AP, L=a + (n-1)d..; L is the last term. In this case: 100

100=2+(n-1)2
n=50

Similarly for odd terms n = 50

Using ur method again, we get

n(n+1)-n^2
=>50 x 51 - 50^2
=>50(51-50)
=>50 x 1
=> 50

Hence option D

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What is the difference between the sum of the even integers from 1–100

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What is the difference between the sum of the even integers from 1–100

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