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by theachiever » Mon Nov 19, 2012 11:47 pm
For positive integer n,the sum of the first n positive integers equals n(n+1)/2.What is the sum of all the even integers between 99 and 301?

A.10,100
B.20,200
C.22,650
D.40,200
E.45,150
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by Bill@VeritasPrep » Mon Nov 19, 2012 11:53 pm
We have a set of even numbers from 100 to 300 (since 99 and 301 are odd), so the range is 200. This means that we have (200/2) + 1 = 101 even numbers. (We add one because it is inclusive).

We also know that the median in an evenly distributed set can be found by finding the mean of the minimum and maximum values. 100 + 300 = 400; 400/2 = 200.

The final step is to multiply 101 and 200. 200 * 101 = 20,200.
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by GMATGuruNY » Tue Nov 20, 2012 3:51 am
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by Brent@GMATPrepNow » Tue Nov 20, 2012 8:49 am
theachiever wrote:For positive integer n,the sum of the first n positive integers equals n(n+1)/2.What is the sum of all the even integers between 99 and 301?

A.10,100
B.20,200
C.22,650
D.40,200
E.45,150
Here's another approach:

We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are 101 integers from 50 to 150 inclusive (150-50+1=101)

To evaluate 2(50+51+52+...+149+150) I'll add values in pairs:

....50 + 51 + 52 +...+ 149 + 150
+150+ 149+ 148+...+ 51 + 50
...200+ 200+ 200+...+ 200 + 200

How many 200's do we have in the new sum? There are 101 altogether.
101x200 = 20,200 = B

Cheers,
Brent
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by Brent@GMATPrepNow » Tue Nov 20, 2012 8:51 am
theachiever wrote:For positive integer n,the sum of the first n positive integers equals n(n+1)/2.What is the sum of all the even integers between 99 and 301?

A.10,100
B.20,200
C.22,650
D.40,200
E.45,150
Alternatively, if we want to evaluate 2(50+51+52+...+149+150), we can evaluate the sum 50+51+52+...+149+150, and then double it.

Important: notice that 50+51+.....149+150 = (sum of 1 to 150) - (sum of 1 to 49)

Now we use the formula:
sum of 1 to 150 = 150(151)/2 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = 20,200 = B

Cheers,
Brent
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