factor26 wrote: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

OG ANSWER IS B PLEASE HELP!

The formula for the sum of numbers is: (# of integers) * (average of first and last term)

To find the number of integers the formula is: (last number - first number) / (difference in pairs) + 1


In this case the average of our first and last number is 100 + 300 (round up from 99 and round down from 301 to make it easier) / 2, therefore our average is 400/2 =200

Now we have to find the number of integers: (300 - 100) / (2) (<--we use 2 because the question wants the number of even integers, and 2 is the difference between all even integers) + 1

Therefore, 200/2 + 1 = 100+1 = 101

Now plug it all in to the original formula:

101*200 = 20 200


I hope this helps !!

Method 1:

Find the sum of the EVEN integers from 99 to 301 is same as Find the sum of the EVEN integers from 100 to 300 both Inclusive.
Find the average of the set (First Term + Last Term)/2 (100+300)/2=200
Find the number of terms (Last Term - First Term)/2 + 1 (300-100)/2 +1= 101
Multiple the average by the number of terms to get answer : 200*101=20,200

Method 2:

The series 100,102,.....300 are in AP. Let the number of terms be n, then the
nth term = first term + ((n-1)* common difference of successive members) = 300 = 100 + (n-1)*2 => n-1 = 100 => n = 101
Sum of all the terms in an AP is equal to 0.5*n*(first term + Last term) = 0.5*101*400 = 20200

Method 3:

Let s = 100+102+104+....300 = 2*(50+51+....150) = 2*(Sum of first 150 positive integers - Sum of first 49 positive integers) = 2*((0.5*150*151)-(0.5*49*50)) = 20200

As Spider man says - 'You always have a choice' and here you have many to choose from !

IMO B
You can rephrase the question as what's the sum of multiples of 2 between 99 and 301.

First and the last even terms in the sequence are 100 and 300

Now, Sum = Avg* # of terms

Avg= (First+Last term)/2 =400/2 = 200
# of terms = (Last - First term)/(multiple of the #) +1 =(300-100)/2 +1 =101
So, the sum is 200*101 = 20200

I have a question need help

Why can't we use sum of 1-301 minus sum of 1-99, then we will get 301*302/2-99*100/2=40501, why it is wrong?

Jonathan - The answer is incorrect because the question is

What is the sum of all the even integers between 99 and 301 ?

and NOT

What is the sum of all the integers between 99 and 301 ?