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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  [#permalink]

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New post 23 Jan 2019, 02:12
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

70% (01:24) correct 30% (01:19) wrong based on 57 sessions

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Re: What is the difference between the sum of the even integers from 1–100  [#permalink]

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New post 23 Jan 2019, 06:11
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
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Re: What is the difference between the sum of the even integers from 1–100  [#permalink]

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New post 24 Jan 2019, 09:12
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

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Ans: D
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Re: What is the difference between the sum of the even integers from 1–100  [#permalink]

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New post 27 Jan 2019, 20:13
Bunuel wrote:
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


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
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Re: What is the difference between the sum of the even integers from 1–100  [#permalink]

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New post 27 Jan 2019, 20:45
Bunuel wrote:
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


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
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Re: What is the difference between the sum of the even integers from 1–100  [#permalink]

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New post 17 Apr 2019, 07:36
Bunuel wrote:
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


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?
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Re: What is the difference between the sum of the even integers from 1–100  [#permalink]

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New post 18 Apr 2019, 15:02
Ish1996 wrote:
Bunuel wrote:
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


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?


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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Re: What is the difference between the sum of the even integers from 1–100   [#permalink] 18 Apr 2019, 15:02
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