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!
For any positive integer n, the sum of
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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 !!
- neelgandham
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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.Method 1:
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
The series 100,102,.....300 are in AP. Let the number of terms be n, then theMethod 2:
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
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)) = 20200Method 3:
As Spider man says - 'You always have a choice' and here you have many to choose from !
Anil Gandham
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- smackmartine
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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
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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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?
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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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 ?
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 ?
Anil Gandham
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Here's one approach.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
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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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
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