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

A. 10,100
B. 20,200
C. 22,650
D. 40,200
E. 45,150

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

a = first term = 100
l = last term = 300

Total number of terms = ((300-100)/2 ) + 1 = 101

sum = n (a + l ) / 2 = 101 (100 + 300 ) / 2 = 20200

sum of all even integer 1 to 301 = 2 * (1+2+...+150) = (2*150*151)/2 = 150*151
sum of all even integer 1 to 99 = 2 * (1+2+....+49) = (2*49*50)/2 = 49*50

required sum = 150*151 - 49*50 = 50*(453 - 49) =
404 * 50 = 20200

thanks!! it makes sense now...what about if the problem was changed to ask for the sum of odd integers? can we amend the formula to get it? thanks!!

Sum = 100+102+...+300
= 100*101 + (0+2+4+...+200) [Taking 100 from each term in series, there are 101 terms]
= 100*101 + 2*(1+2+..+100) [Taking 2 common]
= 100*101 + 2*100*101/2 [Using formula given]
= 2*100*101
= 20200

Answer is (b)
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The sum of n consecutive odd integers starting from 1 is \(n^2\)

e.g. sum of 1 + 3 + 5 + 7 = \(4^2\)

GMAT doesn't expect you to know it but it is a standard formula and doesn't hurt to know.
It can be easily derived using n(n+1)/2 formula. Let's wait for a day or two to see if someone comes up with a simple method of doing it.



We need to find the sum
50 + 51 + 52 +...+ 150 (The sum we need includes 50)

It can be done in two ways:

1. Using n(n + 1)/2

1 + 2 + 3 + ...49 + 50 + 51 + ....150 = 150.151/2
Also, 1 + 2 + 3 + ....+ 49 = 49.50/2

To get the required sum, we just subtract second equation from first.
We get: 50 + 51 + ...150 = 150.151/2 - 49.50/2

2. Using average of AP concept.
Note 50, 51...150 is an arithmetic progression. (Common difference between adjacent terms.)
The average of this AP = (first term + last term)/2 = (50 + 150)/2 = 100
Sum of AP = Average * No of terms = 100 * 101
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There is an easier way (without using formula) :

number of even integers from 99 and 301 = (300-100)/2 + 1 = 101
Average = (100+300)/2 = 200

Sum of all even integers between 99 and 301 = 101 * 200 = 20,200

Question: even integers between 99 and 301

We know 99 is not even and 301 is not even; so ignore both of those

The actual sequence becomes:

100,102,104,106,108,.......,296,298,300

In the above sequence; the first number 100 is divisible by 2(or is even) and the last number 300 is also divisible by 2(or is even).

The common difference "d" is 2.

For such sequences; there is a formula to find the number of elements:

It is {(Last Number - First Number)/d} + 1

So; the above sequence actually contains {(300-200)/2}+1 = 101 elements

And Average = (first number + last number)/2 = (100+300)/2 = 200

Sum = Average * Number of Elements = 101 * 200 = 20200

Ans: 20200
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The sum of the first n terms of an arithmetic pogression is : n/2 *(2a + (n-1)d)

where a is the first term and d is the difference between consecutive terms.


we are to find out the sum of all even integers between 99 and 301

The even numbers look like 100,102,104....300

a=100
n=101
d=2

placing these values in the formula n/2 *(2a + (n-1)d)

sum of even integers = 101/2 * (2*100 + (101-1)*2)
=20200

Choice B.




Alternative method:


The sum of first n positive integers is n(n+1)/2

The sum of first 301 integers is 301*302/2=301*151=45451
The sum of first 99 integers is 99*100/2=99*50=4950

The sum of integers between 99 and 301 = 45451-4950=40501

The above contains all odd and even integers between 99 and 301 --- odd + even = 40501 ------->eqn 1



consider the following series:

numbers in series number of odd numbers sum of odd numbers(so) sum of even numbers(se) so-se
123 2 4 2 2
12345 3 9 6 3
1234567 4 16 12 4


You can see that in the first n integers, the difference of the sum of the odd numbers - sum of even numbers = number of odd numbers in the set.

between 1 and 301 we have 151 odd numbers
between 1 and 99 we have 50 odd numbers

difference of sum of odd numbers and sum of even numbers between 99 and 301 = 151-50=101 ----------> eqn 2

consider eqn 1 and eqn 2

odd+even=40501
odd-even=101

subtract 2 from 1

2even=40400
sum of even numbers between 99 and 301=20200

Choice B

Approach 3

you know that sum of odd + even between 99 and 301 is 40501 (same from previous approach)

The answer should be roughly half of this number --- 20250
closest is B.


Would love to see a simpler more elegant solution. It will save time for us.
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In this question, they give you a formula to reference: the sum of the first n positive integers is n(n+1)/2

But wait, the question is about the sum of EVEN integers, not all positive integers.

We can understand the sum of even integers between 99 and 301 by somehow rewriting in terms of the expression they gave us. They were using the terminology of the first n positive integers.

OK, well sum of even integers between 99 and 301 is the same as between 100 and 300 - inclusive.

Same as the sum of the even integers up to 300 - sum of the even integers up to 98
Same as the sum of the first 150 even integers - sum of the first 49 even integers

So how do we translate this to the expression we were given? We need to translate our "even" integer expression into "positive integers" expression with our formula.

Sum of first 150 even integers = Sum of first 150 POSITIVE integers * 2
1, 2, 3, 4, 5 ===> 2, 4, 6, 8, 10 etc

Sum of first 49 even integers = Sum of first 49 POSITIVE integers * 2

2 * (n(n+1)/2) - 2 * (n(n+1)/2)

Where n=150 in the first case, and n = 49 in the second case

= 2* (150(151)/2) - 2 * (49(50)/2)
= 150*151 - 49*50
= 50 (3*151 - 49)
= 50 (453 - 49)
= 50 (404)
= 50 (400 + 4)
= 20,000 + 200
= 20,200