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For any positive integer n, the sum of

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by gmatpup » Sat Nov 19, 2011 4:24 pm
factor26 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

OG ANSWER IS B PLEASE HELP!

The formula for the sum of numbers is: (# of integers) * (average of first and last term)

To find the number of integers the formula is: (last number - first number) / (difference in pairs) + 1


In this case the average of our first and last number is 100 + 300 (round up from 99 and round down from 301 to make it easier) / 2, therefore our average is 400/2 =200

Now we have to find the number of integers: (300 - 100) / (2) (<--we use 2 because the question wants the number of even integers, and 2 is the difference between all even integers) + 1

Therefore, 200/2 + 1 = 100+1 = 101

Now plug it all in to the original formula:

101*200 = 20 200


I hope this helps :)!!
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by neelgandham » Sat Nov 19, 2011 4:30 pm
Method 1:
Find the sum of the EVEN integers from 99 to 301 is same as Find the sum of the EVEN integers from 100 to 300 both Inclusive.
Find the average of the set (First Term + Last Term)/2 (100+300)/2=200
Find the number of terms (Last Term - First Term)/2 + 1 (300-100)/2 +1= 101
Multiple the average by the number of terms to get answer : 200*101=20,200
Method 2:
The series 100,102,.....300 are in AP. Let the number of terms be n, then the
nth term = first term + ((n-1)* common difference of successive members) = 300 = 100 + (n-1)*2 => n-1 = 100 => n = 101
Sum of all the terms in an AP is equal to 0.5*n*(first term + Last term) = 0.5*101*400 = 20200
Method 3:
Let s = 100+102+104+....300 = 2*(50+51+....150) = 2*(Sum of first 150 positive integers - Sum of first 49 positive integers) = 2*((0.5*150*151)-(0.5*49*50)) = 20200

As Spider man says - 'You always have a choice' and here you have many to choose from !
Anil Gandham
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by smackmartine » Sat Nov 19, 2011 4:31 pm
IMO B
You can rephrase the question as what's the sum of multiples of 2 between 99 and 301.

First and the last even terms in the sequence are 100 and 300

Now, Sum = Avg* # of terms

Avg= (First+Last term)/2 =400/2 = 200
# of terms = (Last - First term)/(multiple of the #) +1 =(300-100)/2 +1 =101
So, the sum is 200*101 = 20200
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by jonathanjing » Sun Sep 16, 2012 9:31 pm
I have a question need help

Why can't we use sum of 1-301 minus sum of 1-99, then we will get 301*302/2-99*100/2=40501, why it is wrong?
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by neelgandham » Mon Sep 17, 2012 6:26 am
Jonathan - The answer is incorrect because the question is

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

and NOT

What is the sum of all the integers between 99 and 301 ?
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by Brent@GMATPrepNow » Mon Sep 17, 2012 7:18 am
factor26 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 (since 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 =20,200 = B

Cheers,
Brent
Last edited by Brent@GMATPrepNow on Mon Sep 17, 2012 7:19 am, edited 1 time in total.
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by Brent@GMATPrepNow » Mon Sep 17, 2012 7:19 am
Alternatively, if we want to evaluate 2(50+51+52+...+149+150), we can 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 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) = 20,200

Cheers,
Brent
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