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
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
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
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
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
