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Integer formula problem

by Sak32 » Sun Nov 10, 2013 2:37 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

B is the answer.
Please provide clear explanations for the solution of this problem.
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by Uva@90 » Sun Nov 10, 2013 3:10 am
Sak32 wrote: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

B is the answer.
Please provide clear explanations for the solution of this problem.
Hi Sak32


You Need to find the Sum of Even numbers between 99 and 301
That is 100+102+104+......296+298+300
=> 2(50+51+.....149+150)

Sum of 51 to 150 can be calculated as sum of (0-150) - (0-49)

As it is given,the sum of the first n positive integers equals (n(n+1))/2

So 0-150 is (150*151)/2 =>(3*50*151)/2
0-49 is (49*50)/2
So above Highlighted statement can be written as

2{50/2(3*151-49)} = 20,200.

Hence answer is [spoiler]B[/spoiler]

Hope it helps you.

Regards,
Uva.
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by GMATGuruNY » Sun Nov 10, 2013 4:00 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 theCodeToGMAT » Sun Nov 10, 2013 4:13 am
Using Direct Formula:

Sum of first EVEN Numbers = (n)(n+1)

Sum from 2 to 300 = (150)(151)

sum from 2 to 98 = (49)(50)

So,

(150)(151) - (49)(50)

50 (3 * 151 - 49)
50 (453 - 49)

50(404)
20200
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by Brent@GMATPrepNow » Sun Nov 10, 2013 8:49 am
Sak32 wrote: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 one more 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), let's 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
Brent Hanneson - Creator of GMATPrepNow.com
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