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
#172 in 13th edition OG
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Approach #1For 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 (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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Approach #2: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
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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Approach #3: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
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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Hi All ,
Can you please correct me , i was doing like below.
n= (100,102,104.......300)
No. of value will be largest - smallest +1
so 300-100+1
201 = No . of value
First term is 100 and the last term is 300
Sum = (first term + last term)(no. of value)/2
so (100+300)(201)/2
which is equal to 40200.
Please correct me and advise me.
Thanks ,
Shreyans
Can you please correct me , i was doing like below.
n= (100,102,104.......300)
No. of value will be largest - smallest +1
so 300-100+1
201 = No . of value
First term is 100 and the last term is 300
Sum = (first term + last term)(no. of value)/2
so (100+300)(201)/2
which is equal to 40200.
Please correct me and advise me.
Thanks ,
Shreyans
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Your calculations for the number of terms (in green) are incorrect.j_shreyans wrote:Hi All ,
Can you please correct me , i was doing like below.
n= (100,102,104.......300)
No. of value will be largest - smallest +1
so 300-100+1
201 = No . of value
First term is 100 and the last term is 300
Sum = (first term + last term)(no. of value)/2
so (100+300)(201)/2
which is equal to 40200.
Please correct me and advise me.
Thanks ,
Shreyans
That formula applies only to consecutive integers (e.g., 4, 5, 6, 7, 8...)
Cheers,
Brent
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Brent@GMATPrepNow wrote:Approach #1For 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 (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
What happened to the factor of 2 around the sum ?
Nevermind, your writing numbers all the way past 100 threw me off - one stops at 100 in actually doing it this way