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by talaangoshtari » Fri Jul 03, 2015 10:01 pm
First we should find the number of the even numbers between 99 and 301. The formula is:
((last - first)/interval) + 1

(300 - 100)/2 + 1 = 101

The sum is equal to
101(100 + 300)/2 = 20200
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by theCEO » Sat Jul 04, 2015 6:57 am
tapanmittal wrote:For any positive integer n,the sum of the first n positive numbers equals n(n+1)/2.What is the sum of all the even integers between 99 and 301?

Ans-20,200
We want the sum of the even numbers from 100 to 300 inclusive
This is equivalent to 2 x (sum of integers from 50 to 150)

2 x (sum of integers from 50 to 150)
2 x (sum of integers from 1 to 150 - sum of integers from 1 to 49)
2 x ( 150*151/2 - 49*50/2)
150*151 - 49*50
50(3*151 - 49)
50(453 - 49)
50(404) = 20200
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by Brent@GMATPrepNow » Sat Jul 04, 2015 8:17 am
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 » Sat Jul 04, 2015 8:17 am
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 » Sat Jul 04, 2015 8:17 am
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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by Jeff@TargetTestPrep » Wed Jul 08, 2015 9:12 am
tapanmittal wrote:For any positive integer n,the sum of the first n positive numbers equals n(n+1)/2.What is the sum of all the even integers between 99 and 301?

Ans-20,200
For any positive integer n, the sum of the first n positive integers equals [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

Solution:

Although a formula is provided in this problem, we can easily solve it using a different formula:

sum = (average)(quantity)

Let's first determine the average.

In any set of numbers in an arithmetic sequence, we can determine the average using the formula:

(1st number in set + last number in set)/2

Remember, we must average the first even integer in the set and the last even integer in the set, which are 100 and 300, respectively. So we have:

(100 + 300)/2 = 400/2 = 200

Next we have to determine the quantity. Once again, we include the first even integer in the set and the last even integer in the set. Thus, we are actually determining the quantity of even consecutive integers from 100 to 300, inclusive.

Two key points to recognize:

1) Because we are determining only the number of "even integers" in the set, we must divide by 2 after subtracting our quantities.

2) Because we are counting the consecutive even integers from 100 to 300, inclusive, we must "add 1" after doing the subtraction (because we are including both 100 and 300 in the calculation).

quantity = [(300 - 100)/2] + 1

quantity = (200/2) + 1

quantity = 101

Finally, we can determine the sum.

sum = 200 x 101

sum = 20,200

Answer:B

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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