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

Hi,

Hoping you can explain this one to me.

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

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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It's crystal clear now. Thank you sir!
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Hi,

Why cant we use the formula, S=n/2(2a+(n-1)d) for this sum. The series is in arthematic progression right?
101/2(2(100)+(100)*2). This would give me a different answer.

lll ask a stupid question
but whats the purpose of the formula in question itself?

if there is no need in it?

The way I see it the formula is there just to make the question a bit harder. An evil trick, if you want.

The OG explanation is equally confusing and I personally do not think is worth the effort to try and master it, there are much "cleaner" ways to solve this problem, as many users here have shown.
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Andy

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 ODD integers between 99 and 301?
A) 10100
B) 20200
C) 20402
D) 40200
E) 45150

Looking for the most efficient way of doing this.

Hi neerajeai, please note that the n(n+1)/2 shortcut formula is only applicable if the starting point is 1. Anytime you want to find the number of terms between two given numbers you should use the general formula ((first - last) / frequency) + 1. You can also multiply by the average at the end to get the sum.

In your case it is (((301-99)/2) + 1) * 200 = 102 * 200 = 20400. Answer choice C with a presumed typo in the unit digit?

Hope this helps!
-Ron
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Ron Awad
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