Kulbushan takes every 3rd number counting from 1001 up to 4000 (1003, 1006, 1009, … , 3997, 4000) and writes them down in sequence. What will be the 251st digit (from the left) in this sequence?

Hi Kumar The question asks for 251st digit, we know that all the numbers in the given range are 4 digit numbers 248/4 =62. So we know that the required digit will be 3rd digit of the 63rd term in the given sequence. The given sequence is an AP with common difference 3, so 63rd term will be
1001 + 62*3 = 1001 +186 = 1187.
So the required digit is 8.
Hope that helps.