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The sum of the first 50 positive odd integers is 2,500.

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by varun289 » Sat Dec 08, 2012 12:45 am
The sum of the first 50 positive odd integers is 2,500. What is the sum of the odd integers from 101 to 199, inclusive?
(A) 4,950
(B) 5,000
(C) 7,450
(D) 7,500
(E) 9,950
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by GMATGuruNY » Sat Dec 08, 2012 3:51 am
varun289 wrote:The sum of the first 50 positive odd integers is 2,500. What is the sum of the odd integers from 101 to 199, inclusive?
(A) 4,950
(B) 5,000
(C) 7,450
(D) 7,500
(E) 9,950
To calculate the sum 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 successive terms.
Since we're adding only the ODD integers here, the interval is 2.
Thus, the number of odd integers from 101 to 199 = (199-101)/2 + 1 = 50.
Average of biggest and smallest = (199+101)/2 = 150.
Sum = (number of integers) * (average of biggest and smallest) = 50*150 = 7500.

The correct answer is D.
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by Brent@GMATPrepNow » Sat Dec 08, 2012 9:07 am
varun289 wrote:The sum of the first 50 positive odd integers is 2,500. What is the sum of the odd integers from 101 to 199, inclusive?
(A) 4,950
(B) 5,000
(C) 7,450
(D) 7,500
(E) 9,950
Here's another approach:

The sum of the first 50 positive odd integers is 2,500
So, 1+3+5+7+...+97+99=2500

We want to find the sum of 101+103+105+...+197+199
Notice that we can take the above sum and rewrite it as:
100+1 + 100+3 + 100+5 +....100+97 + 100+99
Rearrange to get: 100+100+100+...+100+100 + 1+3+5+7+...+97+99
This equals (50)(100) + 2500
Which equals 5000 + 2500 = [spoiler]7500 = D[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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