Let's solve this problem without using any mathematical formula of Arithmetic Progression. This gives us the speedy answer which is so required in the competitive exams where there is a constraint of time.
The greatest three digit number is 999.
And 999/19 = 52.57.
This implies upto 999, a total of 52 numbers are completely divisible.
Subtracting from this, number of two- digit numbers also divisible by 19 taken into account in the previous calculation, i.e 19,38,57,76, and 95, we get :
52-5 = 47
Thus there are 47 such 3 digit numbers in all.
There should be 3 digit number,So First Digit=114
Last digit which is divisible by 19 is 19*52=988
Apply the formula of A.P....An=a+(n-1)d---------------Eq(1)
Where An=988, a=114, common difference d= 19
Applying above value in Eq(1) we get,