20*(20+1)/2 = 210
The sum of the first 20 integers is 190.
There are 20.
Two of them. Two of them. Two of them. Two of them.
The integers 2 and 10 have a product of 20 and a sum of 12.
To find the sum of the series of the first 20 positive integers ending in 3, we first need to identify the pattern. The series would be 3, 13, 23, 33, ..., 193. This is an arithmetic series with a common difference of 10. To find the sum, we can use the formula for the sum of an arithmetic series: n/2 * (first term + last term), where n is the number of terms. Plugging in the values, we get 20/2 * (3 + 193) = 10 * 196 = 1960. Therefore, the sum of the series is 1960.
The sum of the first 20 integers is 190.
The sum of the first four non-negative, consecutive, even integers is 20.
There are 20.
The sum of all integers from 1 to 20 inclusive is 210.
Many people would consider 0 to be the first even positive integer, however similar to the neutron in an atom, 0 holds a neutral position. Meaning it is neither a proton (+1) or electron (-1). So our first positive even integer would be 2.2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110, which is the sum of the first ten evenly positive integers.
Two of them. Two of them. Two of them. Two of them.
To find the last two digits of the sum of the factorials of the first 100 positive integers, we can observe that for ( n \geq 10 ), ( n! ) ends with at least two zeros due to the factors of 10 in the factorial (from the pairs of 2 and 5). Therefore, we only need to calculate the sum of the factorials from 1! to 9!. The sum ( 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! ) equals 40320, and the last two digits are 20. Thus, the last two digits in the sum of factorials of the first 100 positive integers are 20.
The integers are 18 and 20.
The integers 2 and 10 have a product of 20 and a sum of 12.
To find the sum of the series of the first 20 positive integers ending in 3, we first need to identify the pattern. The series would be 3, 13, 23, 33, ..., 193. This is an arithmetic series with a common difference of 10. To find the sum, we can use the formula for the sum of an arithmetic series: n/2 * (first term + last term), where n is the number of terms. Plugging in the values, we get 20/2 * (3 + 193) = 10 * 196 = 1960. Therefore, the sum of the series is 1960.
The integers are 19, 20 and 21.
20,21 and 22 are three consecutive integers whose sum is 63. 20+21+22