Q: What is the sum of the first ten even numbers?

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It is 110.

thousands

100

-55

No, it is impossible.

The first ten positive numbers total 55.

The sum of the first 10 natural numbers is 51.

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

To find the sum of the first ten prime numbers, you first have to find what the first 10 prime numbers are. They are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. Added together they equal 129.

to print the sum of first ten numbers:- void main() {int i,sum; sum=0; while(i<=10) sum=sum+i; i=i+1; } printf("The sum is : %d",sum); }

There is only 1 even prime number which is 2

The sum first Prime number is (2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 ) = 129 The sum of the first composite numbers is (4 + 6 + 8 + 9 + 10 + 12 + 14 + 15 + 16 + 18) = 112 The difference is 129 - 112 = 17 !!!!!

Five of them.

The sum of the first ten positive integers is: 55

Sum = 0 For N = 1 to 10 Sum = Sum + 2*N Next N Print Sum

The sum of the first ten positive integers, i.e. 1,2,3,4,5,6,7,8,9, and 10, is 55. The sum of the first ten negative integers, i.e. -1,-2,-3,-4,-5,-6,-7,-8,-9, and -10 is -55. The sum of the first ten positive integers plus the sum of the first ten negative integers is 0.

Two is the ONLY even prime number.

There is only one even prime number. It is the number two. All other even numbers are composite.

double[] numsToSum; // the numbers you want to find the sum of double sum = 0.0; for(int i = 0; i < numsToSum.length; ++i) { sum += numsToSum[i]; }

The sum is 110.The numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20.EquationFor the first n numbers that are multiples of 2, the sum is (n)(n+1).If zero is included as a "whole number multiple of 2," then the sum is (n)(n-1).

#include#includevoid main(){int i,sum=0;for(i=1;i

$n = 10*(1+10)/2;

pls ans this

#include<conio.h> #include<stdio.h> void main() { float num,avg,sum=0; int count=1; clrscr(); printf("Enter ten numbers: \n"); while(count<=10) { scanf("%d",&num); sum+=num; count++; } avg=sum/--count; printf("\n\nThe sum of the ten numbers = %d",sum); printf("\nThe average of the ten numbers = %d",avg); getch(); }

When the sum of the numbers is ten or greater.

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