Problem Solving

the sum of the first 50 positve even integer is 2,550. What is the sum of the even integers from 102 to 200, inclusive? If anyone can help, I'd really appreciate it...

sum of the even integers from 102 to 200, inclusive means we can use the basic formula of AP. Sn = sum of n digits = n/2(2a + (n - 1)d)
Where a = first term of the series, d = common diff and n = no of terms.
Tn = nth term = a + (n - 1)d = 102 + (n - 1) * 2 = 200 or n = 50
So Sn = 50/2(102 + 200). Hope u get me, mrkylee!

camitava,

you are correct and the answer comes out to 7550

camitava/Tsonam,

why is it that the nth term equation is equated to 200? is this because of the inclusive aspect?


thanks

Bschool08, u r correct! As the Qs is saying that 200 is the max limit - inclusive. So we will take 200 as the nth term. Hope u understand what I want to mean!

yes i do- thanks Camitava!!

100+ 102 .... 200

= 2+ 4+6.... +100 + 50(100)

= 2550 + 50*100 = 7550