fifty-five
1+2+3+4+5 = 15
6+7+8+9+10 = 40
15+40 = 55
Wiki User
∙ 9y ago1 1 1
11
main() { sum=0; float avg=0.0; int sqr[10]; for(i=1;i<=10;i++) { sqr[i]=(i*i); } for(i=1;i<=10;i++) { sum=sum+sqr[i]; } avg=sum/10; dts it !!!
The numbers are 11 and 10.
16
The sum of the first 10 counting numbers (1-10) is 51.
The composite numbers between 1 and 10 are 4, 6, 8, 9, 10. And their sum is 37.
The two numbers 10 and -1: 10 × -1 = -10 10 + -1 = 10 - 1 = 9
11
It is 10*(10+1)/2 = 55
10
The general equation to find the sum of the numbers 1 to n is: (n*(n+1))/2So, for n=10, you have:(10*(10+1))/2(10*11)/2110/255
sum = 0 for(n = 0; n <= 10; n += 2) sum += n;
There are many ways to get a sum of 10. Here are some examples: -2+6+2=10 -8+1+1=10 -3+6+1=10 -1+3+6=10 -0+1+9=10
The sum of the first 10 even numbers is 110.
The sum is 420
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); }