it that answer true?

Yes, the answer is correct.
what is the query?

Sum of Set = n[(last-first)/2) where n = [(last-first)/k]+1 and the value of K= the distance between numbers in a set.

Step 1: Find K.

The question is a sum of all multiples of 3, which would mean 3,6,9,12... the distance between each is (6-3)=3 or (9-6)=3 and so on. As such, the value of k for this problem is 3.

Step 2: determine the first and last values that fit the parameters of the question.

While we are given a range of 250 to 350, the first and last values are not 250 and 350 becuase they are NOT multiples of 3. The shorthand way to determine if something is a multiple of 3 is to add up all the digits. If the number you get from doing that is a number divisible by 3, then the larger original number is ALSO divisible by 3.

EX: 3312 is divisible by 3 because 3+3+1+2 =9 and 9 is clearly divisible by 3. Similarly, 23451321 is ALSO divisible by 3 because 2+3+4+5+1+3+2+1 = 21, a number you know is divisible by 3.

Anyway, 250 = 7, which is not divisible by 3 and 350 = 8 which is not divisible by 3. By testing numbers one by one above 250 and below 350 (within the range) we see that the first multiple of 3 is actually 251 and the last multiple of 3 in our range is 348.

Step 3: Solve for n

n = [(348-252)/3]+1 .... this is equal to (96/3)+1, which becomes 32+1 = 33

Step 4: With our n value in tow, we can now solve the original equation.

sum of set = n[(first+last)/2] = 33*[(252+348)/2] = 33*[300] = 9900.

Step 5: Compare Quantity A and B

A = 9900. B = 9990. Therefore, B is greater than A and the answer is B

Another approach

The first number divisible by \(3\) after \(250\) is \(252\) Like: 252= 84 * 3
The first number divisible by \(3\) before \(350\) is \(348\) Like: 348 = 116 * 3
Hence there are altogether so the range means the total number here is, 84........116 = ( 116-84+1 )=33
Since the 33 numbers are equally spread so, 33/2 * (252 + 348) = 9900
option B > option A

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