2 n in a d is 2 = nickels in a dime.
The sum of an arithmetical sequence whose nth term is U(n) = a + (n-1)*d is S(n) = 1/2*n*[2a + (n-1)d] or 1/2*n(a + l) where l is the last term in the sequence.
Assuming nickles, dimes, and quarters, there are ten different ways to make change for a half dollar. Just enumerate the combinations... 10 n 8 n 1 d 6 n 2 d 4 n 3 d 2 n 4 d 5 d 5 n 1 q 3 n 1 d 1 q 1 n 2 d 1 q 2 q
1 q 1 d 1 q 2 n 2 d 3 n 3 d 1 n 1 d 5 n 7 n
This can be found out by using the formula for sum of n terms of an arithmetic progression. here, n is not known. a=2 d=2 (since it's even) nth term=98 (last even number less than 100) using formula for nth term, nth term=a+(n-1)d 98=2+(n-1)2 therefore, n=49 so, sum of n terms=n/2[2*a+(n-1)d] putting n=49, a=2, and d=2, sum=2450
Base (b) = 1 000 000 Number (n) = 2 416 304 Modular (m) =n Mod b = 416304 Integer Difference (d) = n - b = 2 if m/b is more than or equal to 0.5 then d = d+1 else d answer is (d), 2 million
Pick one 1: for (i=2*n-1; i>=1; i-=2) printf ("%d\n", i); 2: for (i=n; i>=1; i--) printf ("%d\n", 2*i-1); 3: for (i=0; i<n; i++) printf ("%d\n", 2*(n-i)-1);
Sum = n/2[2Xa1+(n-1)d] where n is last number, a1 is the first number & d is the common difference between the numbers, here d=2 for the even /odd numbers. Sum = n/2 [2Xa1+(n-1)2]
There are 22 ways to make change from a dollar using nickels, dimes, and quarters. 1. 4 q 2. 10 d 3. 20 n 4. 2 q , 5 d 5. 3 q , 2 d , 1 n 6. 1 q , 7 d, 1 n 7. 9 d, 2 n 8. 8 d, 4 n 9. 7 d, 6 n 10. 6 d , 8 n 11. 5 d , 10 n 12. 4 d , 12 n 13. 2 d , 16 n 14. 1 d , 18 n 15. 5 n , 3 q 16. 3 n , 1 q , 6 d 17. 7 n , 1 q , 4 d 18. 9 n , 1 q , 3 d 19. 11 n , 1 q , 2 d 20. 13 n , 1 q , 1 d 21. 14n , 3 d 22. 15n , 1 q
The sum of an arithmetical sequence whose nth term is U(n) = a + (n-1)*d is S(n) = 1/2*n*[2a + (n-1)d] or 1/2*n(a + l) where l is the last term in the sequence.
Assuming nickles, dimes, and quarters, there are ten different ways to make change for a half dollar. Just enumerate the combinations... 10 n 8 n 1 d 6 n 2 d 4 n 3 d 2 n 4 d 5 d 5 n 1 q 3 n 1 d 1 q 1 n 2 d 1 q 2 q
d = n + 7 and (n + 17)/(d - 6) = 2 so (n + 17)/ n + 7 - 6 = 2 (n + 17)/(n + 1) = 2 2n + 2 = n + 17 n = 15 and d = 22 Check n + 17 = 32, d - 6 = 16, 32/16 = 2 Original number is 15/22
1 q 1 d 1 q 2 n 2 d 3 n 3 d 1 n 1 d 5 n 7 n
In a sequence of numbers, a(1), a(2), a(3), ... , a(n), a(n+1), ... he first differences are a(2) - a(1), a(3) - a(2), ... , a(n+1) - a(n) , ... Alternatively, d the sequence of first differences is given by d(n) = a(n+1) - a(n), n = 1, 2, 3, ...
A fraction, n/d, in percentage terms is simply (100*n) / d. In this case, n = 2 and d = 3.
int n, N;N = some even numberfor (n=2; n
// transpose for the sparse matrix void main() { clrscr(); int a[10][10],b[10][10]; int m,n,p,q,t,col; int i,j; printf("enter the no of row and columns :\n"); scanf("%d %d",&m,&n); // assigning the value of matrix for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("a[%d][%d]= ",i,j); scanf("%d",&a[i][j]); } } printf("\n\n"); //displaying the matrix printf("\n\nThe matrix is :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("%d\t",a[i][j]); } printf("\n"); } t=0; printf("\n\nthe non zero value matrix are :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { // accepting only non zero value if(a[i][j]!=0) { t=t+1; b[t][1]=i; b[t][2]=j; b[t][3]=a[i][j]; } } } printf("a[0 %d %d %d\n",m,n,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,b[i][1],b[i][2],b[i][3]); } a[0][1]=n; a[0][2]=m; a[0][3]=t; int s[10],u[10]; if(t>0) { for(i=1;i<=n;i++) { s[i]=0; } for(i=1;i<=t;i++) { s[b[i][2]]=s[b[i][2]]+1; } u[1]=1; for(i=2;i<=n;i++) { u[i]=u[i-1]+s[i-1]; } for(i=1;i<=t;i++) { j=u[b[i][2]]; a[j][1]=b[i][2]; a[j][2]=b[i][1]; a[j][3]=b[i][3]; u[b[i][2]]=j+1; } } printf("\n\n the fast transpose matrix \n\n"); printf("a[0 %d %d %d\n",n,m,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,a[i][1],a[i][2],a[i][3]); } getch(); }
d = 2(n = 265782341)