All even integers between 101 and 999 are 102, 104, 106, 108, . . . . 998.

First term(a) = 102, common difference(d) = 104 – 102 = 2 and nth term(a_n) = 998.

By using the formula of nth term of an A.P.

a_n = a + (n – 1)d

So, 

=> 998 = 102 + (n – 1)2

=> 2(n – 1) = 896

=> n – 1 = 448

=> n = 449

Now by using the formula of sum of n terms of an A.P.

S_n = n[a + a_n] / 2

So, 

S₄₄₉ = 449[102 + 998] / 2

= 449[550]

= 246950