gwiz87 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
Here's one approach. We want 100+102+104+....298+300
This equals 2(50+51+52+...+149+150)
From here, a quick way is to evaluate this is to first recognize that there are
101 integers from 50 to 150 inclusive (150 - 50 + 1 =
101)
To evaluate 2(50+51+52+...+149+150), let's add values in pairs:
....50 + 51 + 52 +...+ 149 + 150
+
150+ 149+ 148+...+ 51 + 50...200+ 200+ 200+...+ 200 + 200
How many 200's do we have in the new sum? There are
101 altogether.
101 x 200 = 20,200
Answer: B
Approach #2: From my last post, we can see that we have 101 even integers from 100 to 300 inclusive.
Since the values in the set are
equally spaced, the average (mean) of the 101 numbers = (first number + last number)/2 = (100 + 300)/2 = 400/2 = 200
So, we have 101 integers, whose average value is 200.
So, the sum of all 101 integers = (101)(200)
= 20,200
= B
Approach #3: Take 100+102+104+ ...+298+300 and factor out the 2 to get 2(50+51+52+...+149+150)
From here, we'll evaluate the sum
50+51+52+...+149+150, and then double it.
Important: notice that
50+51+.....149+150 =
(sum of 1 to 150) -
(sum of 1 to 49) Now we use the given formula:
sum of 1 to 150 = 150(151)/2 = 11,325sum of 1 to 49 = 49(50)/2 = 1,225So,
sum of 50 to 150 =
11,325 -
1,225 =
10,100So, 2(50+51+52+...+149+150) = 2(10,100) = 20,200
Answer: B
Cheers,
Brent
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