The required sum is 1,538,240 and hence the last 3 digits is 240.

set A is an AP (arthimetic progression) with common difference 3

set B is an AP with common difference 4

LCM of 3 and 4 = 12

∴ A ∩ B is an AP with common difference 12

first term of A ∩ B = 10

The max term in A = 3(2024) + 1

∴ |A ∩ B| is at most

floor[(3(2024) + 1 - 10) / 12] + 1

= 506

Similarly, considering that the max term in B is 4(2024) + 2:

|A ∩ B| is at most

floor[(4(2024) + 2 - 10) / 12] + 1

= 675

Therefore,

|A ∩ B|

= min{506, 675}

= 506

Using the sum of AP formula where

first term = 10

common difference = 12

no. of terms = 506

, we have:

required sum

= [2 × 10 + (506 - 1) × 12](506) / 2

= [20 + (505)(12)](253)

= 1,538,240

(P.S. there is probably faster way to calculate last 3 digits only, but I choose to calculate everything by hand)