Sum of all integers between two integers
The sum of all integers between two integers refers to the total result of adding up all the whole numbers (integers) that are between two given integers, including the two given integers themselves.
For example, suppose we want to find the sum of all integers between 3 and 7, inclusive. The integers between 3 and 7 are 3, 4, 5, 6, and 7. To find their sum, we simply add them up:
3 + 4 + 5 + 6 + 7 = 25
So the sum of all integers between 3 and 7 is 25.
In general, if we want to find the sum of all integers between two integers a and b (inclusive), where a and b are whole numbers and a ≤ b, we can use the formula:
int n = max - min; // Calculated sum int result = ((n + 1) *min) + ((n *(n + 1)) / 2);
min is start number and max is ending number.
Program Solution
/*
C program for
Sum of all integers between two integers
*/
#include <stdio.h>
void sumBetweenAtoB(int min, int max)
{
if (max < min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
int n = max - min;
// Calculated sum
int result = ((n + 1) *min) + ((n *(n + 1)) / 2);
// Display calculated result
printf("\n Sum of number from (%d..%d) is : %d", min, max, result);
}
int main(int argc, char
const *argv[])
{
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
sumBetweenAtoB(55, 70);
return 0;
}
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
/*
Java program for
Sum of all integers between two integers
*/
public class Addition
{
public void sumBetweenAtoB(int min, int max)
{
if (max < min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
int n = max - min;
// Calculated sum
int result = ((n + 1) * min) + ((n * (n + 1)) / 2);
// Display calculated result
System.out.print("\n Sum of number from (" +
min + ".." + max + ") is : " + result );
}
public static void main(String[] args)
{
Addition task = new Addition();
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
task.sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
task.sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
task.sumBetweenAtoB(55, 70);
}
}
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
// Include header file
#include <iostream>
using namespace std;
/*
C++ program for
Sum of all integers between two integers
*/
class Addition
{
public: void sumBetweenAtoB(int min, int max)
{
if (max < min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
int n = max - min;
// Calculated sum
int result = ((n + 1) *min) + ((n *(n + 1)) / 2);
// Display calculated result
cout << "\n Sum of number from (" << min
<< ".." << max << ") is : " << result;
}
};
int main()
{
Addition *task = new Addition();
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
task->sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
task->sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
task->sumBetweenAtoB(55, 70);
return 0;
}
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
// Include namespace system
using System;
/*
Csharp program for
Sum of all integers between two integers
*/
public class Addition
{
public void sumBetweenAtoB(int min, int max)
{
if (max < min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
int n = max - min;
// Calculated sum
int result = ((n + 1) * min) + ((n * (n + 1)) / 2);
// Display calculated result
Console.Write("\n Sum of number from (" +
min + ".." + max + ") is : " + result);
}
public static void Main(String[] args)
{
Addition task = new Addition();
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
task.sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
task.sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
task.sumBetweenAtoB(55, 70);
}
}
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
package main
import "fmt"
/*
Go program for
Sum of all integers between two integers
*/
func sumBetweenAtoB(min, max int) {
if max < min {
return
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
var n int = max - min
// Calculated sum
var result int = ((n + 1) * min) + ((n * (n + 1)) / 2)
// Display calculated result
fmt.Print("\n Sum of number from (", min, "..", max, ") is : ", result)
}
func main() {
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
sumBetweenAtoB(4, 10)
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
sumBetweenAtoB(-5, 6)
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
sumBetweenAtoB(55, 70)
}
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
// Rust program for
// Sum of all integers between two integers
fn main(){
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
sum_between_ato_b(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
sum_between_ato_b(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
sum_between_ato_b(55, 70);
}
fn sum_between_ato_b(min: i32, max: i32) {
if max < min {
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
let n: i32 = max - min;
// Calculated sum
let result: i32 = ((n + 1) * min) + ((n * (n + 1)) / 2);
// Display calculated result
print!("\n Sum of number from ({}..{}) is : {}", min, max, result);
}
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
<?php
/*
Php program for
Sum of all integers between two integers
*/
class Addition
{
public function sumBetweenAtoB($min, $max)
{
if ($max < $min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
$n = $max - $min;
// Calculated sum
$result = (($n + 1) * $min) + ((int)(($n * ($n + 1)) / 2));
// Display calculated result
echo("\n Sum of number from (".$min.
"..".$max.
") is : ".$result);
}
}
function main()
{
$task = new Addition();
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
$task->sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
$task->sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
$task->sumBetweenAtoB(55, 70);
}
main();
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
/*
Node JS program for
Sum of all integers between two integers
*/
class Addition
{
sumBetweenAtoB(min, max)
{
if (max < min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
var n = max - min;
// Calculated sum
var result = ((n + 1) * min) + (parseInt((n * (n + 1)) / 2));
// Display calculated result
process.stdout.write("\n Sum of number from (" +
min + ".." + max + ") is : " + result);
}
}
function main()
{
var task = new Addition();
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
task.sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
task.sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
task.sumBetweenAtoB(55, 70);
}
main();
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
# Python 3 program for
# Sum of all integers between two integers
class Addition :
def sumBetweenAtoB(self, min, max) :
if (max < min) :
return
# Formula
# resutl = ((n+1) *min) + ((n *(n+1))/2))
# // Here
# min : starting number
# max : last number
# n = max - min
# n indicate number of elements between min to max
# ((n *(n+1))/2) sum of natural number from 1 to n
n = max - min
# Calculated sum
result = ((n + 1) * min) + (int((n * (n + 1)) / 2))
# Display calculated result
print("\n Sum of number from (",
min ,"..", max ,") is : ", result, end = "")
def main() :
task = Addition()
# Test A
# min = 4
# max = 10
# [4+5+6+7+8+9+10] = 49
task.sumBetweenAtoB(4, 10)
# Test B
# min = -5
# max = 6
# [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
task.sumBetweenAtoB(-5, 6)
# Test C
# min = 55
# max = 70
# [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
task.sumBetweenAtoB(55, 70)
if __name__ == "__main__": main()
Output
Sum of number from ( 4 .. 10 ) is : 49
Sum of number from ( -5 .. 6 ) is : 6
Sum of number from ( 55 .. 70 ) is : 1000
# Ruby program for
# Sum of all integers between two integers
class Addition
def sumBetweenAtoB(min, max)
if (max < min)
return
end
# Formula
# resutl = ((n+1) *min) + ((n *(n+1))/2))
# // Here
# min : starting number
# max : last number
# n = max - min
# n indicate number of elements between min to max
# ((n *(n+1))/2) sum of natural number from 1 to n
n = max - min
# Calculated sum
result = ((n + 1) * min) + ((n * (n + 1)) / 2)
# Display calculated result
print("\n Sum of number from (",
min ,"..", max ,") is : ", result)
end
end
def main()
task = Addition.new()
# Test A
# min = 4
# max = 10
# [4+5+6+7+8+9+10] = 49
task.sumBetweenAtoB(4, 10)
# Test B
# min = -5
# max = 6
# [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
task.sumBetweenAtoB(-5, 6)
# Test C
# min = 55
# max = 70
# [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
task.sumBetweenAtoB(55, 70)
end
main()
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
/*
Scala program for
Sum of all integers between two integers
*/
class Addition()
{
def sumBetweenAtoB(min: Int, max: Int): Unit = {
if (max < min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
var n: Int = max - min;
// Calculated sum
var result: Int = ((n + 1) * min) + ((n * (n + 1)) / 2);
// Display calculated result
print("\n Sum of number from (" +
min + ".." + max + ") is : " + result);
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: Addition = new Addition();
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
task.sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
task.sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
task.sumBetweenAtoB(55, 70);
}
}
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
/*
Swift 4 program for
Sum of all integers between two integers
*/
class Addition
{
func sumBetweenAtoB(_ min: Int, _ max: Int)
{
if (max < min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
let n: Int = max - min;
// Calculated sum
let result: Int = ((n + 1) * min) + ((n * (n + 1)) / 2);
// Display calculated result
print("\n Sum of number from (",
min ,"..", max ,") is : ",
result, terminator: "");
}
}
func main()
{
let task: Addition = Addition();
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
task.sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
task.sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
task.sumBetweenAtoB(55, 70);
}
main();
Output
Sum of number from ( 4 .. 10 ) is : 49
Sum of number from ( -5 .. 6 ) is : 6
Sum of number from ( 55 .. 70 ) is : 1000
/*
Kotlin program for
Sum of all integers between two integers
*/
class Addition
{
fun sumBetweenAtoB(min: Int, max: Int): Unit
{
if (max < min)
{
return;
}
/*
Formula
resutl = ((n+1) *min) + ((n *(n+1))/2))
// Here
min : starting number
max : last number
n = max - min
n indicate number of elements between min to max
((n *(n+1))/2) sum of natural number from 1 to n
*/
val n: Int = max - min;
// Calculated sum
val result: Int = ((n + 1) * min) + ((n * (n + 1)) / 2);
// Display calculated result
print("\n Sum of number from (" +
min + ".." + max + ") is : " + result);
}
}
fun main(args: Array < String > ): Unit
{
val task: Addition = Addition();
// Test A
/*
min = 4
max = 10
[4+5+6+7+8+9+10] = 49
*/
task.sumBetweenAtoB(4, 10);
// Test B
/*
min = -5
max = 6
[-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
*/
task.sumBetweenAtoB(-5, 6);
// Test C
/*
min = 55
max = 70
[55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
*/
task.sumBetweenAtoB(55, 70);
}
Output
Sum of number from (4..10) is : 49
Sum of number from (-5..6) is : 6
Sum of number from (55..70) is : 1000
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