27+n=n+27
n/3 + 27
To find the sum of the numbers, we must first know the value of n. This was not included in your question.
To find the sum of the first 27 odd numbers, we can use the formula for the sum of an arithmetic series: Sn = n^2, where n is the number of terms. In this case, n = 27, so the sum is 27^2 = 729. Therefore, the sum of the first 27 odd numbers is 729.
n + (n + 1) + (n + 2) = 27 3n + 3 = 27 n = 8 Thus, the numbers are 8, 9, and 10, and the least of these is 8.
12
The two equal numbers that have a sum of 54 are 27 and 27. When you add these two numbers together, you get 54. This is because the sum of two equal numbers will always be double the value of one of the numbers. In this case, 27 + 27 = 54.
let the number be n, then the sum of the number and 2 is n+2
If the number is n, then twice the number is 2n and the sum of the two is n + 2n = 3n.
The sum of the interior angles of a polygon with n sides is (n-2)*180 degrees. Here, n = 27, so the sum of the interior angles is (27-2)*180 = 25*180 = 4500 degrees.
n = number n + 6 ======
This is simple algebra. Let the unknown number be n. Then the sum of this number and 8 = n + 8.
void main () { int no1, sum, n; clrscr() sum = 0; n = 1; printf("\n enter the number to which sum is to be generated"); scanf("%d",&no1); while(n<=no1) { sum=sum+n; n=n+1 } printf("\n the sum of %d = %d, no1, sum"); getch (); }