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PS Question help please!

by dqure040 » Wed Nov 04, 2015 7:52 pm
For any positive integer n, the sum of the first n positive integers equal n(n+10)/2. What is the sum of all even integers between 99 and 301?
a) 10,100 b)20,200 c)22,650 d)40,200 e)45,150

Could use some help here.
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by theCEO » Wed Nov 04, 2015 8:27 pm
dqure040 wrote:For any positive integer n, the sum of the first n positive integers equal n(n+1)/2. What is the sum of all even integers between 99 and 301?
a) 10,100 b)20,200 c)22,650 d)40,200 e)45,150

Could use some help here.
Sum of even numbers from 99 to 301 can be written as:
100+102+104+106...+300
=2 x (50+51+52....+150)
=2 x (sum of the first 150 numbers - sum of the first 49 numbers)
=2 x (150*151/2 - 49x50/2)
=(150*151) - (49*50) = 22650 - 2450 = 20200
ans = b
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by Brent@GMATPrepNow » Wed Nov 04, 2015 8:44 pm
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
Approach #1

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) I'll 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.
101x200 = [spoiler]20,200 = B[/spoiler]

Cheers,
Brent
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by Brent@GMATPrepNow » Wed Nov 04, 2015 8:44 pm
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
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

Cheers,
Brent
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by Brent@GMATPrepNow » Wed Nov 04, 2015 8:45 pm
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
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 = 11,325
sum of 1 to 49 = 49(50)/2 = 1,225

So, sum of 50 to 150 = 11,325 - 1,225 = 10,100

So, 2(50+51+52+...+149+150) = 2(10,100) = [spoiler]20,200 = B[/spoiler]

Cheers,
Brent
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