For any positive integer n, the sum of the first n positive integers

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.