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by [email protected] » Sun Apr 14, 2013 3:13 am
Please show me a simple solution to this, thanks
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by Anju@Gurome » Sun Apr 14, 2013 3:25 am
Please try to post the problem in words so that later people can search for it.

Anyway, have a look at this post >> https://www.beatthegmat.com/og13-q172-ps ... tml#624461
Three different methods to solve this problem has been discussed there.
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Quant Expert, Gurome

Backup Methods : General guide on plugging, estimation etc.
Wavy Curve Method : Solving complex inequalities in a matter of seconds.

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by GMATGuruNY » Sun Apr 14, 2013 4:09 am
For any positive integer n, the sum of first n positive integer 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
The question asks for the sum of the even integers from 100 to 300, inclusive.

Ignore the formula given. Instead, the following can be used to calculate the sum of any set of evenly spaced integers:

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)/interval + 1

The INTERVAL is the distance between one term and the next.
Since we're adding only the even integers here, the interval 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.
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by Brent@GMATPrepNow » Sun Apr 14, 2013 4:36 am
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 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.
101 x 200 = [spoiler]20,200 = B[/spoiler]

Cheers,
Brent
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by Brent@GMATPrepNow » Sun Apr 14, 2013 4:40 am
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


Another approach uses the formula provided in the question.
We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
To 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
Brent Hanneson - Creator of GMATPrepNow.com
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