even integers

2 + 4 + 6 + ... + 100 = 2550
Now, 102 + 104 + ... 200
= (2 + 100) + (4 + 100) + ... (100 + 100)
= (2 + 4 + ... 100) + (100 + 100 ... 50 times)
= 2550 + 100*50
= 2550 + 5000
= 7550

There is another way of solving such problems, for that please remember this formula:-
(N/2)((2 x A) + (N-1) D)
Where:-
N - Total number of digits you have to add
A - The first digit
D - Difference between two digits

Now we'll tackle the problem
Here we know that the first digit is 102 so A = 102
Difference between two digits is 2, as all are even so D = 2
Total number of even digits between 102 and 200 are 50 so N = 50

Putting the values in the formula we get
(50/2) ((2 x 102) + (49) 2)
=25 (204 + 98)
=25 (302)
=7550 (The ans)
With this formula you can calculate the sum of all the APs

Hope this helps!

Good one Vinay, AP would be a good & generic way to solve this one unless you can think of

2550 + 50*100 =7550

In the formula for AP that Vinay has given ,I think he meant Nos instead of digits

(N/2)((2 x A) + (N-1) D)
Where:-
N - Total number of Numbers you have to add
A - The first or starting Number in the sequence
D - Difference between two Numbers (common or same)

Although first statement is irrelevant here. But I liked givemeanid's approach. Thanks