Sum of even integers

[Da_Gr8_Mperor]

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

[GmatJunky]

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

 

In question ignore the formula, only there to confuse...

see attachment

[nenuphani]

sum of first 49 cons. integers = 49 * 50 /2 = 49*25 = A

sum of first 150 cons. integers = 150 * 151/2 = 151 * 75 = B

 

so, sum of even no.s b/w 99 & 301 ==> 100 + 102 + ... + 300

==> 2(50+51+42 + ... + 150)

==> 2(B-A) = 2(151*75-49*25) = 20200

[jmeasp]

sum of n even no=n(n+1)

No of even no from 1 to 301=301/2=150

sum of even no upto 301=150x151

 

No of even no from 1 to 99=99/2=49

sum of even no upto 99=49x50

 

therefore sum of even nos from 99 to 301=150x151-49x50

=20200

[manish8109]

a=100 , d=300 , n=300-100/2 + 1 = 101

 

S = (100+300)101/2 = 200*101 = 20200(ANS)

[gmatlove]

S = 100+102+.....300

a= 100

d = 2

n=101

 

Sum of Series = (n/2)(2a+(n-1)d)

 

101/2(200+100*2)

101*200

20200

 

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