//sum and product of 3 nos #include #include void main() { int a,b,c; printf("enter the 3 nos"); scanf("%d%d%d",&a,&b,&c); printf("sum of 3 nos",a+b+c); printf("product of 3 nos",a*b*c); getch(); }
The nos. Are 8 and -8 Sum of the nos. 8+(-8)=0 Product of the nos. 8 × (-8) = -64
148
Each one of them can be expressed as a sum of two primes.
11x11
777/3 = 259. So the numbers are 258, 259 and 260.
The Nth partial sum is the sum of the first n terms in an infinite series.
2 and 3.
if (n%2==0) sum=n/2*(n+1); else sum=(n+1)/2*n;
5 + 7 = 12
i dnt know i hoped that helped u
Find the Sum to n terms of the series 5 5+55+555+ +n Terms