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

[kamalj]

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


OAB

Can you please check my solution and let me know where i am getting wrong.

# of terms = (300-100)/2 + 1 = 101

Sum = n(n+1)/2

101(102)/2

101X51

Doing this unit digit will be 1, which is not in the options.

Please explain.

[Brent@GMATPrepNow]

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


OAB

Can you please check my solution and let me know where i am getting wrong.

# of terms = (300-100)/2 + 1 = 101

Sum = n(n+1)/2

101(102)/2

101X51

Doing this unit digit will be 1, which is not in the options.

Please explain.

You are correct to say that there are 101 terms in the sequence.
However, the formula you used (sum = n(n+1)/2) does not necessarily apply here.
That formula tells us the sum of the first n CONSECUTIVE integers.
That is, 1 + 2 + 3 + 4 + 5 + . . . + n = n(n+1)/2

So, for example, 1 + 2 + 3 + 4 + 5 + . . . + 101 = 101(101+1)/2
BUT, this is not what you are asked to find.
You are asked to find the sum of 100 + 102 + 104 + 106 + . . . . 298 + 300 (not the sum 1 + 2 + 3 + 4 + 5 + . . . + 101)

Does that help?

Cheers,
Brent

[Brent@GMATPrepNow]

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

[Brent@GMATPrepNow]

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

[Brent@GMATPrepNow]

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

[Scott@TargetTestPrep]

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

We are asked to find the sum of the even integers from 100 to 300 inclusive. We can use the following formula (rather than the one that has been provided):

Sum = quantity x average

The quantity is the number of even integers from 100 to 300, inclusive, which is:

(300 - 100)/2 + 1 = 101

The average of the even integers from 100 to 300, inclusive, is also the average of 100 and 300, which is:

(100 + 300)/2 = 200

Thus, the sum of the even integers from 100 to 300, inclusive, is:

Sum = 101 x 200

Sum = 20,200

Answer: B