Sum of even integers !

[Uva@90]

For any positive integers 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

Answer: B

Thanks in advance.

[Brent@GMATPrepNow]

For any positive integers 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

Answer: B

Thanks in advance.

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 = B

Cheers,
Brent

[Brent@GMATPrepNow]

For any positive integers 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

Answer: B

Thanks in advance.

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 $$$$

[Brent@GMATPrepNow]

For any positive integers 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

Answer: B

Thanks in advance.

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

[Uva@90]

For any positive integers 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

Answer: B

Thanks in advance.

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

Brent : Thanks a Ton.
You took your time to explaining me all the possible ways to solve the problem.
I thank you once again.
Regards,
Uva.

[send2dar]

Thanks Brent.. I felt approach 2 is the quickest... I liked approach 1 too..

Regards
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"Trying for 750+"