Find the sum of all even integers between 101 and 999.

In this problem, we need to find the sum of all the even numbers lying between 101 and 999.

So, we know that the first even number after 101 is 102 and the last even number before 999 is 998.

Also, all these terms will form an A.P. with the common difference of 2.

So here,

First term (a) = 102

Last term (l) = 998

Common difference (d) = 2

So, here the first step is to find the total number of terms. Let us take the number of terms as n.

Now, as we know,

$a_{n}=a+(n-1) d$

So, for the last term,

$998=102+(n-1) 2$

$998=102+2 n-2$

$998=100+2 n$

$998-100=2 n$

Further simplifying,

$898=2 n$

$n=\frac{898}{2}$

$n=449$

Now, using the formula for the sum of n terms,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

For n = 64, we get,

$S_{n}=\frac{449}{2}[2(102)+(449-1) 2]$

$=\frac{449}{2}[204+(448) 2]$

$=\frac{449}{2}(204+896)$

$=\frac{449}{2}(1100)$

On further simplification, we get,

$S_{n}=449(550)$

$=246950$