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Find N Arithmetic Means between A and B
  • Difficulty Level : Medium
  • Last Updated : 28 Nov, 2019

Given three integers A, B and N the task is to find N Arithmetic means between A and B. We basically need to insert N terms in an Arithmetic progression. where A and B are first and last terms.

Examples:


Input : A = 20 B = 32 N = 5
Output : 22 24 26 28 30
The Arithmetic progression series as 
20 22 24 26 28 30 32 

Input : A = 5  B = 35  N = 5
Output : 10 15 20 25 30

Approach :
Let A1, A2, A3, A4……An be N Arithmetic Means between two given numbers A and B . Then A, A1, A2 ….. An, B will be in Arithmetic Progression .
Now B = (N+2)th term of the Arithmetic progression .
So :
Finding the (N+2)th term of the Arithmetic progression Series
where d is the Common Difference

B = A + (N + 2 – 1)d
B – A = (N + 1)d

So the Common Difference d is given by.



d = (B – A) / (N + 1)

So now we have the value of A and the value of the common difference(d),
now we can find all the N Arithmetic Means between A and B.

C++




// C++ program to find n arithmetic 
// means between A and B
#include <bits/stdc++.h>
using namespace std;
   
// Prints N arithmetic means between
// A and B.
void printAMeans(int A, int B, int N)
{
    // calculate common difference(d)
    float d = (float)(B - A) / (N + 1);
       
    // for finding N the arithmetic 
    // mean between A and B
    for (int i = 1; i <= N; i++) 
        cout << (A + i * d) <<" ";    
}
   
// Driver code to test above 
int main()
{
    int A = 20, B = 32, N = 5;
    printAMeans(A, B, N);    
    return 0;
}

Java




// java program to illustrate
// n arithmetic mean between 
// A and B
import java.io.*;
import java.lang.*;
import java.util.*;
   
public class GFG {
   
    // insert function for calculating the means
    static void printAMeans(int A, int B, int N)
    {       
        // Finding the value of d Common difference
        float d = (float)(B - A) / (N + 1);
                              
        // for finding N the Arithmetic 
        // mean between A and B
        for (int i = 1; i <= N; i++) 
          System.out.print((A + i * d) + " ");
           
    }
   
    // Driver code
    public static void main(String args[])
    {
        int A = 20, B = 32, N = 5;
        printAMeans(A, B, N);
    }
}

Python3




# Python3 program to find n arithmetic
# means between A and B
  
# Prints N arithmetic means 
# between A and B.
def printAMeans(A, B, N):
  
    # Calculate common difference(d)
    d = (B - A) / (N + 1)
      
    # For finding N the arithmetic 
    # mean between A and B
    for i in range(1, N + 1): 
        print(int(A + i * d), end = " "
  
# Driver code
A = 20; B = 32; N = 5
printAMeans(A, B, N) 
  
# This code is contributed by Smitha Dinesh Semwal

C#




// C# program to illustrate
// n arithmetic mean between 
// A and B
using System;
  
public class GFG {
  
    // insert function for calculating the means
    static void printAMeans(int A, int B, int N)
    {     
        // Finding the value of d Common difference
        float d = (float)(B - A) / (N + 1);
                              
        // for finding N the Arithmetic 
        // mean between A and B
        for (int i = 1; i <= N; i++) 
        Console.Write((A + i * d) + " ");
          
    }
  
    // Driver code
    public static void Main()
    {
        int A = 20, B = 32, N = 5;
        printAMeans(A, B, N);
    }
}
// Contributed by vt_m

PHP




<?php
// PHP program to find n arithmetic 
// means between A and B
  
// Prints N arithmetic means 
// between A and B.
function printAMeans($A, $B, $N)
{
      
    // calculate common
    // difference(d)
    $d = ($B - $A) / ($N + 1);
      
    // for finding N the arithmetic 
    // mean between A and B
    for ($i = 1; $i <= $N; $i++) 
        echo ($A + $i * $d) ," "
}
  
    // Driver Code 
    $A = 20; $B = 32; 
    $N = 5;
    printAMeans($A, $B, $N);
      
// This code is Contributed by vt_m.
?>
Output:
22 24 26 28 30

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