Sum of even integer

[sodha.rakesh]

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2.
What is the sum of all the even integer between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

[GMATGuruNY]

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2.
What is the sum of all the even integer between 99 and 301?

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

We need to count the even integers from 100 to 300, inclusive.

Ignore the formula given. Instead, use the following:

Sum = (number of integers) * (average of biggest and smallest)

To count the number of evenly spaced integers in a set:

Number of integers = (Biggest - Smallest)/(distance between each successive pair) + 1

Since we're adding only the even integers, the distance between each successive pair is 2.
Thus, the number of even integers from 100 to 300 = (300-100)/2 + 1 = 101.
Average of biggest and smallest = (300+100)/2 = 200.
Sum = (number of integers) * (average of biggest and smallest) = 101*200 = 20,200.

The correct answer is B.

[blaster]

the simple logic of this question is you should find all the numers sum from 1...301, also find 1..99 then the first result minus the second. the number you will get will be around 40000 . but in question we have additional restriction. the even numbers. as in a number line we have one odd and one even you need just divide the result to 2.

[Anurag@Gurome]

the simple logic of this question is you should find all the numers sum from 1...301, also find 1..99 then the first result minus the second. the number you will get will be around 40000 . but in question we have additional restriction. the even numbers. as in a number line we have one odd and one even you need just divide the result to 2.

Not so simple!
Just consider the integers from 1 to 10.
Sum of all integers = 55
Sum of odd integers = 25 ≠ (55/2)
Sum of even integers = 30 ≠ (55/2)

In fact in this case following your method will result a sum equal to 20,250.5, which is not correct.

If you have to use the given formula, then it can be done as follows,

• .. Sum of all the even integer between 99 and 301
  = (100 + 102 + 104 + ... + 300)
  = 2*(50 + 51 + 52 + ... + 150)
  = 2*[(1 + 2 + 3 + ... + 150) - (1 + 2 + 3 + ... + 49)]
  = 2*[(150*151)/2 - (49*50)/2]
  = [(150*151) - (49*50)]
  = 50*(3*151 - 49)
  = 50*404
  = 20,200

The solution provided by Mitch is more efficient and time saving.

[hovhannest]

Arithmetic progression... sum = (a1+an)*n/2, for our case sum = (100+300)*n/2, where n=101 (note please that between [100 - 300] 100 odd and 101 even numbers).

Thus, 200*101=20,200