Answer: A
A: The numbers of multiples of 3 between 102 and 729:
We should find the minimum and maximum multiples of 3 in this range of numbers.
102 is a multiple of 3 and the maximum multiple of 3 in this range is 729.
Multiples of 3: 102, 105, 108, …., 720, 723, 726, 729
How many numbers are there in this sequence? (Max - min)/ distance +1 = (729 - 102) / 3 + 1
As we see A is 1 value bigger than B and is the answer.
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series formed will be:-
102,105......., 729
an = a+ (n-1)*d
729 = 102 + (n-1)*3
729-102
------- + 1 = n
3
Hence A is greater by 1

I would take a small example to answer this.

3-9.

3-9 have 3 multiples of 3

9-3 / 3 = 2

So multiples inclusive of numbers will be 1 value greater than range by the number.

This is valid for any range. Hence the answer

This is an evenly spaced set so it is an arithmetic series also.
Now the nth term in the arithmetic series is given by:
an = a1 + (n-1)d
where n is the total no of terms in the series and d is the difference between the two consecutive terms.
We have an=729, a1=102 and d=3 as they are the multiples of 3.
So plugging in the values we get
729 = 102 + (n-1)3
After simplifying we get the A=B+1
So A is the answer.