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Answer: 45
Three, six, nine, twelve, and fifteen are the first five multiples of three.
The sum of 3 + 6 + 9 + 12 + 15 equals 45.
As a result, the total of the first five multiples of three equals 45.
3, 6, 9, 12, 15 are other viable options.
Here, n = 5, a = 3, and l = 15 are the values.
The sum of AP with the first term a and the last term l is:
(n/2)(a + l) = S
(3 + 15) = (5/2)
= (18/5) *(5/2)= 5*9=45
As a result, the required amount is 45.
