Q: What is the sum of the first 100 positive multiples of 4?

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The sum of the first 100 positive even numbers is 10,100.

Sum of the first 15 positive integers is 15*(15+1)/2 = 120 Sum of the first 15 multiples of 8 is 8*120 = 960

The sum of the first 100 odd numbers (1 through 199) is 10000 (ten thou)

The sum of the first 25 numbers is 25*26/2 = 325 So the sum of the first 25 multiples of 12 is 325*12 = 3900

5 + 10+15+20=50 ans

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The sum of the first 100 positive even numbers is 10,100.

Sum of the first 15 positive integers is 15*(15+1)/2 = 120 Sum of the first 15 multiples of 8 is 8*120 = 960

The sum of the first 100 positive even numbers can be calculated using the formula for the sum of an arithmetic series: n*(first term + last term)/2. In this case, the first term is 2, the last term is 200, and n is 100. Therefore, the sum is 10,100.

10100.

The sum of the first 10 multiples of 3 is 165.

The sum of the first 100 odd numbers (1 through 199) is 10000 (ten thou)

We have to assume that you're talking about whole numbers. The sum is 5,050 .

Their sum is 1200.

Sum_ap = ½ × number_of_terms × (first_term + last_term) For the first 88 multiples of three: number_of_terms = 88 first_term = 1 × 3 = 3 last_term = 88 × 3 = 264 → Sum = ½ × 88 × (3 + 264) = 44 × 267 = 11748

69342

They are 13.

The sum of the first 25 numbers is 25*26/2 = 325 So the sum of the first 25 multiples of 12 is 325*12 = 3900