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by Rahul@gurome » Sat Jan 15, 2011 8:45 pm
We have to find 102 + 104 + 106 +....+ 200 = (100 + 2) + (100 + 4) + (100 + 6) +....+ (100 + 100)
= (100 + 100 + 100 +...+ 100) + (2 + 4 + 6 +...+ 100)

It is given that 2 + 4 + 6 + .... + 100 = 2,550

Now 100 will occur 50 times, so the above sum can be written = (50 × 100) + 2,550 = 5,000 + 2,550 = 7,550

[spoiler]So, the correct answer is B.[/spoiler]
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by ankur.agrawal » Sat Jan 15, 2011 9:44 pm
No of even integers from 102---200 = [(200-102)/2]+1 = 50 even integers.

Sum of the even integers from 102 to 200 inclusive = Average of even integers* no of even integers.

Average of Even Integers from 102- 200 inclusive = (200+102)/2=151. (applying the formula for consecutive integers)

Therefore: Sum= 151*50=7550

The Answer is B.
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by GMATGuruNY » Sun Jan 16, 2011 3:52 am
trangle wrote:The sum of the first 50 positive even integers is 2,550. What is the sum of the even integers from 102 to 200 inclusive?

a. 5,100
b. 7,550
c. 10,100
d. 15,500
e. 20,100

OA: B
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)/(distance between each successive pair) + 1

The distance between each successive pair of even integers is 2.
Thus, the number of even integers from 102 to 200 = (200-102)/2 + 1 = 50.
Average of biggest and smallest = (200+102)/2 = 151.
Sum = (number of integers) * (average of biggest and smallest) = 50*151 = 7550.

The correct answer is B.
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by gmat7202011 » Sun Jan 16, 2011 7:54 am
What is the sum of the even integers from 102 to 200 inclusive?

The number of even terms / integers between 102 and 200 is [200-102]/2 + 1 = 49

This is an AP with a common difference , say variable a, of 2, with the first term as 102

You can use the formula

Sum = n/2[ 2a + (n-1)d]

n=49, a =102, d =2

Subsitute to get the desired answer.
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by samrat_bagchi » Mon Jan 17, 2011 10:16 am
gmat7202011 wrote:What is the sum of the even integers from 102 to 200 inclusive?

The number of even terms / integers between 102 and 200 is [200-102]/2 + 1 = 49

This is an AP with a common difference , say variable a, of 2, with the first term as 102

You can use the formula

Sum = n/2[ 2a + (n-1)d]

n=49, a =102, d =2

Subsitute to get the desired answer.
the no. of even terms shd be 50..
the solution process is however correct.
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