The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
Can we solve this with the help of pattern recognition
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The sum of the first 50 positive even integers is 2550.[email protected] wrote:The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
In other words, 2 + 4 + 6 + 8 + ...+ 98 + 100 = 2550
We want the sum: 102 + 104 + 106 + . . . 198 + 200
IMPORTANT: Notice that each term in this sum is 100 greater than each term in the first sum.
In other words, 102 + 104 + 106 + . . . 198 + 200 is the SAME AS...
(100 + 2) + (100 + 4) + (100 + 6) + ... + (100 + 98) + (100 + 100)
We can rearrange these terms to get: (100 + 100 + ... + 100 + 100) + (2 + 4 + 6 + 8 + ...+ 98 + 100)
IMPORTANT: There are 50 100's in the red sum, and we're told that the blue sum = 2550
So, our sum = 50(100) + 2550
= 5000 + 2550
= 7550
= B
Cheers,
Brent
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Another approach:[email protected] wrote:The sum of the first 50 positive even integers is 2550. What is the sum of even integers from 102 to 200 inclusive?
A. 5100
B. 7550
C. 10100
D. 15500
E. 20100
We want to evaluate: 102 + 104 + 106 + . . . 198 + 200
Since the terms in this series are EQUALLY SPACED, the sum = (average of first and last term)(# of terms)
So, sum = [(102 + 300)/2][50]
= [(302)/2][50]
= [151][50]
= 7550
= B
Cheers,
Brent
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Hi shibsriz,
There's a variation on Brent's approach called "bunching", which can also be used to answer this question.
Since we're dealing with the EVEN integers from 102 to 200, we have a total of 50 numbers
Adding the biggest (200) and smallest (102) gives us 302
Adding the next biggest (198) and next smallest (104) also gives us 302
With 50 numbers, we have 25 groups of two. Each group of two = 302
25(302) is approximately 25(300), which = 7500
There's only one answer that's close to 7500
Final Answer: B
GMAT assassins aren't born, they're made,
Rich
There's a variation on Brent's approach called "bunching", which can also be used to answer this question.
Since we're dealing with the EVEN integers from 102 to 200, we have a total of 50 numbers
Adding the biggest (200) and smallest (102) gives us 302
Adding the next biggest (198) and next smallest (104) also gives us 302
With 50 numbers, we have 25 groups of two. Each group of two = 302
25(302) is approximately 25(300), which = 7500
There's only one answer that's close to 7500
Final Answer: B
GMAT assassins aren't born, they're made,
Rich