To find the sum of the first 100 positive multiples of 4, we can use the formula for the sum of an arithmetic series: Sn = n/2 * (a1 + an), where Sn is the sum, n is the number of terms, a1 is the first term, and an is the last term. In this case, a1 = 4, an = 4*100 = 400, and n = 100. Plugging these values into the formula, we get: Sn = 100/2 * (4 + 400) = 50 * 404 = 20,200. Therefore, the sum of the first 100 positive multiples of 4 is 20,200.
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
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
They are 13.
69342
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