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Consecutive Intergers Problem

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by jzebra10 » Sun Dec 04, 2011 5:54 am
I don't understand exactly what the problem is telling/asking. How do you go about solving this problem?

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all even integers between 99 and 301.

a) 10,100
b) 20,200
c) 22,650
d) 40,200
e) 45,150

To solve problems like these, do we have to use the above formula?

Thanks.
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by user123321 » Sun Dec 04, 2011 6:16 am
jzebra10 wrote:I don't understand exactly what the problem is telling/asking. How do you go about solving this problem?

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all even integers between 99 and 301.

a) 10,100
b) 20,200
c) 22,650
d) 40,200
e) 45,150

To solve problems like these, do we have to use the above formula?

Thanks.
need to find sum of even integers b/w 99 & 301
i.e between 100 & 300 (including both)
100+102+...+298+300 = 2(50+51+...+149+150)
= 2(sum of first 150 numbers - sum of first 49 numbers)
= user above formula and solve above equation you will get B.

another way..
since it is pretty much standard formula you can also use sum of n numbers in a series.
whree series is 100,102,104....298,300
no of terms in this series = (300-100)/2 + 1 = 101
sum of this series = 101(99+301)/2 = 101*200 = B

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by LalaB » Sun Dec 04, 2011 6:26 am
https://www.beatthegmat.com/need-help-wi ... 14127.html
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by GMATGuruNY » Sun Dec 04, 2011 7:14 am
jzebra10 wrote:I don't understand exactly what the problem is telling/asking. How do you go about solving this problem?

For any positive integer n, the sum of the first n positive integers equals n(n+1)/2. What is the sum of all even integers between 99 and 301.

a) 10,100
b) 20,200
c) 22,650
d) 40,200
e) 45,150

To solve problems like these, do we have to use the above formula?

Thanks.
I would ignore the formula given. I posted a solution here:

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