Consider a denominator of r;

It has proper fractions:

1/r, 2/r, ...., (r-1)/r

Their sum is: (1 + 2 + ... + (r-1))/r

The numerator of this sum is

1 + 2 + ... + (r-1)

Which is an Arithmetic Progression (AP) with r-1 terms, and sum:

sum = number_of_term(first + last)/2

= (r-1)(1 + r-1)/2

= (r-1)r/2

So the sum of the proper fractions with a denominator or r is:

sum{r} = ((r-1)r/2)/r = ((r-1)r/2r = (r-1)/2

Now consider the sum of the proper fractions with a denominator r+1:

sum{r+1} = (((r+1)-1)/2

= ((r-1)+1)/2

= (r-1)/2 + 1/2

= sum{r) + 1/2

So the sums of the proper fractions of the denominators forms an AP with a common difference of 1/2

The first denominator possible is r = 2 with sum (2-1)/2 = ½;

The last denominator required is r = 100 with sum (100-1)/2 = 99/2 = 49½;

And there are 100 - 2 + 1 = 99 terms to sum

So the required sum is:

sum = ½ + 1 + 1½ + ... + 49½

= 99(½ + 49½)/2

= 99 × 50/2

= 2475

The sum of the series

a + ar + ar2 + ... is a/(1 - r) for |r| < 1

Sum of r = summer

one fourth the sum of r and ten is identical to r munus 4

void main() { int num,r,sum=0; clrscr(); printf("enter the number\n"); scanf("%d",&num); while(num!=0) { r=num%10; sum=sum+r; num=num/10 } printf("The sum of individual digit of given number is=%d",sum); getch() }

For an Arithmetic Progression,

Sum = 15[a + 7d].{a = first term and d = common difference}

For a Geometric Progression,

Sum = a[1-r^15]/(r-1).{r = common ratio }.

5(r+5) is.

The formula for the sum of the series r(1/n2-1/n2) is r(1-1/n2).

#include<stdio.h>

#define PIE 3.142

void main()

{

int r,sum;

printf("Enter the radius of area\n");

scanf("%d",&r);

sum=PIE*r*r;

printf("The area of circle is:%d",sum);

}

0.0714

this is actually an equation

let the 'sum' be x

we are assuming r means are meaning =

so

x/6 = 7

x=42

so thats the answer

" 5(p+r) " is.

r - 10 + 100 - 3r = -2r + 90

Could be (r + 8)/2 or r/2 + 4

Erm..they r and there not coz some r natural and sum r coz of humans

As an expression it is: 5(p+r) which means 5p+5r

the circumference ... 2 pi r

nt rely dah controls r weird buhh sum r

#include
int main()
{
int num, r, sum=0, temp;
print f("enter a number:");
scan f("%d",&num);
temp=num;
while(n!=0)
{
r=num%10;
num=num/10;
sum=sum*10+r;
}
if(temp==sum)
print f("%d is a palindrome",temp);
else
print f("%d is not a palindrome",temp);
return 0;
}

some are underpaid on work

the answer to the sum of two thirds of 1700 is 1133.33 R

asdfghjkl;

(r+5)(r-4)

The idea is to get two numbers whose product in this case is -20, and whose sum in this case s +1.

nope, none of them r brothers......just best friends :)

in youtube there r sum sngs

. r u thick? or thin? espanol njrgvbjkseg

there r 800 vetices in three triangles

sum are such as jews are cold blooded

Suppose the first term is a, the second is a+r and the nth is a+(n-1)r.

Then the sum of the first five = 5a + 10r = 85

and the sum of the first six = 6a + 15r = 123

Solving these simultaneous equations, a = 3 and r = 7

So the first four terms are: 3, 10, 17 and 24

/*This program takes a square matrix from user and check whether it's a Magic Square or not. */

#include<stdio.h>

#include<conio.h>

int num[10][10],sum[21],r,c,i=0,x;

void read_matrix()

{ for(r=0;r<x;r++)

{ for(c=0;c<x;c++)

{ printf("Enter the value of matrix[%d][%d] :",r,c);

scanf("%d",&num[r][c]);

}

}

}

void calculate_sum()

{ int sum_row,sum_col=0,ctr=0;

for(r=0;r<x;r++)

{ sum_row = 0;

for(c=0;c<x;c++)

{ printf("\t%d",num[r][c]);

sum_row = sum_row + num[r][c];

}

sum[ctr] = sum_row;

ctr++;

printf(":: %d",sum_row);

printf("\n");

}

while(i<x)

{ printf("\t::");

i++;

}

printf("\n");

for(c=0;c<x;c++)

{sum_col = 0;

for(r=0;r<x;r++)

sum_col = sum_col + num[r][c];

sum[ctr] = sum_col;

ctr++;

printf("\t%d",sum_col);

}

for(r=0,c=0;r<x;r++,c++)

sum[2*x] = sum[2*x] + num[r][c];

for(r=0,c=x-1;r<x;r++,c--)

sum[2*x+1] = sum[2*x+1] + num[r][c];

}

char check_matrix()

{ char c;

for(i=0;i<2*x+1;i++)

{ if(sum[i] == sum[i+1])

c = 'Y';

else

{ c = 'N';

break;

}

}

return c;

}

void main()

{ char c;

clrscr();

printf("Enter d no. of rows or cols u want in your matrix :");

scanf("%d",&x);

read_matrix();

calculate_sum();

c = check_matrix();

if(c=='Y')

printf("\nThis matrix is a Magic Square");

else

printf("\nThis matrix is not a Magic Square");

getch();

}

S.I. = (P x R x T)/100

where R is rate, T is time, P is Original sum and S.I. is simple interest.

800x100 = P x R x T

P = 80000/(5x7) = 80000/35 = 2285.71

So, the original sum is Rs 2285.71

since you do not know what r and s is all you need to write is...

2r + 2s

the sum just simply means add them together.

Suppose the nth term is = arn where n = 1,2,3, ...

Then the sum to the nth term is a*(rn+1 - 1)/(r - 1) or, equivalently,

a*(1 - rn+1)/(1 - r)

In honey trees.U can c dat trees r green but sum of dem r yellow.Da yellow 1s r da honey trees.

The sum to infinity of a geometric series is given by the formula S∞=a1/(1-r), where a1 is the first term in the series and r is found by dividing any term by the term immediately before it.

It is the sum of:

1) Initial endowment (inheritance).

2) Life-time wages.

3) Sum of interest on wealth (e-r).

This can be represented mathematically in a form such as:

I = e-r[k0 + int(wAL(t))dt]

Well, it's very hard to write a flowchart in text, so I'll give you some pseudo code instead.

int number = the given number

int sum = 0

loop while number is not 0

sum = sum + (number mod 10)

number = number / 10

your nan minus your mum squared all dived by 2 to the sum of r=20 to r= 1.

this just goes to sho dat sum wurds r uzeless

NPV=NFV/(1+r)^n The role of the "(1+r)^n" is to discount the future money to what it is worth in todays dollars. The 1 accounts to the sum itself and the plus r takes into account the interest rate. NPV=NFV/(1+r)^n The role of the "(1+r)^n" is to discount the future money to what it is worth in todays dollars. The 1 accounts to the sum itself and the plus r takes into account the interest rate.

The sum = 2r : where r is the row number.

#include<stdio.h>

#include<conio.h>

main()

{

int n,s,r,t;

clrscr();

printf("enter n");

scanf("%d",&n);

s=0;t=0;

while(n!=0)

{

r=n%10;

{

if(r%2!=0)

t=t+r;

if(r%2==0)

s=s+r;

}

n=n/10;

}

printf("sum of even position digits%d\n",s);

printf("sum of odd position digits%d\n",t);

getch();

}

The sum of a geometric sequence is a(1-rn)/(1-r)

In this case, a = 8, r = -2 and n=15

So the sum is 8(1-(-2)15)/(1+2)

=8(1+32768)/3

=87,384

So the sum of the first 15 terms of the sequence 8, -16, 32, -64.... is 87,384.

I hav lots of friends sum of them r Frey,Yaz,Beth,Gab.......................

why do you want to know r u gunna kill sum one

.....5

.....Σ 9r

...r=1