OG 12th Q#157

[ketandoshi]

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

Ketan

[Rahul@gurome]

Solution:
We need the sum 100+102+104+........+300 = 2(50+51+52+............+150)
= 2{50 + (50+1) + (50+2).............+(50+100)}
= 2{50*101 + (1+2+3........+100)} = 2{5050 + (100*101)/2} (using the formula that the sum of first n +ve integer equals n(n+1)/2)
= 20200

[clock60]

i wonder for what we are given the n(n-1)/2
it is simple sum of AP, there are 101 even numbers between 99 and 301
so sum=((100+300)/2)*101=20200

[chris@veritasprep]

I agree with clock60! I have always found it strange that the formula n(n-1)/2 is given in this problem. While it can surely be used to solve this question, it is really an arithmetic progression problem that is easier to solve with an understanding of average = sum of terms/ # of terms and thus sum of terms (what we want here) = (average of terms) x (# of terms).

For these types of problems (which I call unusual sum problems - quite common) they are best done by finding the average of the terms and multiplying that by the number of terms. In this problem, we want the sum of all even numbers between 99 and 301 so lets first calculate the average of the set:

In any evenly spaced set such as this, the average of the set is the (first term + last term)/2 Here that is (99+301)/2 = 200.

To complete the problem, we now need the number of terms to multiply with the average. In these problems, it is usually the number of terms that trip people up. How many even numbers are there between 99 and 301?

That is the same question as how many even numbers are there between 100 and 300, inclusive? (the first and last even numbers in that set are 100 and 300) For determining the number of terms in an inclusive set problem, take the difference between the end terms and divide by the spacing of the terms and add one to the result. Here that is (300 - 100)/2 + 1 = 101.

To complete the problem, if the average is 200 and the number of terms is 101 then the sum is 200 x 101 = 20,200

In my opinion, this is an important concept for the GMAT so let me give another question to practice with:

What is the sum of all even integers from 650 to 750, inclusive?

(A) 35,000
(B) 35,700
(C) 70,000
(D) 70,700
(E) 77,000


For the inclusive set calculations here are a couple more drills:

How many multiples of 10 are there between 1000 and 2000, inclusive?

How many even numbers are there between 29 and 101?



Discussion to follow if needed and answers are below:





B, 101, 36

[Baten80]

99 and 301 are not even integers
so renge of even integers will from 100 to 300 i.e., 300-100=200/2+1=101

(300+100)/2*101=20200 Ans.

[edwardsmichele]

Thank you Clock60 and Chris@VeritasPrep for your most helpful replies. Chris, an added thanks for the extra examples you provided which meant I did not have to go searching all over the place for additional practice.

Best regards.

Shelley