did i take the right approach

Anurag@Gurome wrote:

sohailmbaprep wrote:Q) for any +ve integer n,sum of 1st n +ve integers equals n(n+1)/2 ...what is the sum of all even integers between 99 and 301.

----this is OG 13th editions Question number 172---

I solved it like ...
from 99 to 301 there are 202 numbers, and out of them 101 will be even numbers....
so now applying AP formula :- sum=n/2(1st term+last term)
sum = 101/2 (100+300)

though solving this i got the right answer...OG explains the answer in different way...is my approach correct,actually m confused ,i shouldn't have included 99 and 301 ...question says terms between them, not including them.
If i m wrong then pls explain how can i solve this.

Also pls tell how can i know the total number of terms in a series...say how many numbers are there from 505 to 710 and how many numbers are there between 505 and 710

There is a much better way to solve this without using the given formula.

For any set of uniformly spaced integers, the average of the integers = (Largest integer + Smallest integer)/2

In this case, all the even integers between 99 and 300, i.e. from 100 to 300 are uniformly spaced. Hence, average of them = (100 + 300)/2 = 200

Now, sum of all these integers = (Average)*(Number of integers in the range) = 200*[(300 - 100)/2 + 1] = 200*101 = 20,200