Bunuel wrote:
What is the sum of all the even integers between 99 and 401 ?
(A) 10,100
(B) 20,200
(C) 37,750
(D) 40,200
(E) 45,150
TL;DR
Sum of n terms = n*((a1+an)/2)
# of terms = (400 - 100)/2 + 1 = 200 - 50 + 1 = 151
Avg = (400+100)/2 = 250
Sum = 250 * 151 = 37750
ANSWER: C
Veritas Prep Official Solution
We can solve this question in multiple ways.
Required Sum = 100 + 102 + 104 + 106 + … + 400
Method 1:
We know the sum of consecutive even integers but only when they start from 2. So what do we do? We find the sum of even integers starting from 2 till 400 and subtract the sum of even integers starting from 2 till 98 from it! Note that we subtract even numbers till 98 because 100 is a part of our series.
How many even integers are there from 2 to 400? I hope you agree that we will have 200 even integers in this range (both inclusive)
Sum of these 200 integers = 200(201)
How many even integers are there from 2 to 98? Now we have 49 even integers here.
Sum of these 49 even integers = 49(50)
What is the sum of integers from 100 to 400? It will be 200(201) – 49(50) = 40200 – 2450 = 37750
Method 2:
100 + 102 + 104 + 106 + … + 400 = 2( 50 + 51 + 52 + 53 + … + 200) (We take out 2 common and find the sum in brackets)
Sum in brackets = 50 + 51 + 52 + 53 + … + 200
We know the sum of consecutive integers but only when they start from 1. So we find the sum of first 200 numbers and subtract the sum of first 49 numbers from it. That will give us the sum of numbers from 50 to 200. Note that we subtract 49 numbers because 50 is a part of our series.
Sum of 1 + 2 + 3 + … + 200 = 200*201/2 = 20100
Sum of 1 + 2 + 3 + … + 49 = 49*50/2 = 1225 (I am doing these calculations here only for clarity. Normally, I would like to carry all these till the last step, then take common, divide by whatever I can etc so that I have very few actual calculations left.)
Therefore, 50 + 51 + 52 + 53 + … + 200 = 20100 – 1225 = 18875
Then 100 + 102 + 104 + 106 + … + 400 = 2*18875 = 37750
Answer (C).