OG13 Q172 (PS)

[basso25@]

i cannot follow the answer explanation; please break down in simple terms for me. thank you!

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

[mohit_1607]

To Calculate: Sum of all the even integers between 99 and 301
Therefore, we have to calculate the sum of the following AP:
100, 102, 104....,300

Sum of AP = n/2[2a + (n-1)d] = n/2[a + l] (where l is the last term in AP)
Here, n = 101
a = 100
l = 300

Sum = 101/2[100 + 300] = 20200
-> Option B