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.Â
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 Bimport 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 BÂ
# Prints N arithmetic means # between A and B.def printAMeans(A, B, N):Â
    # 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 = " ") Â
# Driver codeA = 20; B = 32; N = 5printAMeans(A, B, N) Â
# 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 BÂ
// 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) ," "; }Â
    // Driver Code     $A = 20; $B = 32;     $N = 5;    printAMeans($A, $B, $N);     // This code is Contributed by vt_m.?> |
Javascript
<script>Â
// JavaScript program to find n arithmetic// means between A and BÂ
// Prints N arithmetic means // between A and B.function printAMeans(A, B, N){Â
    // Calculate common difference(d)    let d = (B - A) / (N + 1)         // For finding N the arithmetic     // mean between A and B    for(let i = 1; i < N + 1; i++)         document.write(Math.floor(A + i * d)," ") }Â
// Driver codelet A = 20, B = 32, N = 5;printAMeans(A, B, N) Â
// This code is contributed by ShinjanpatraÂ
</script> |
Output:
22 24 26 28 30
Â
 Time Complexity : O(N) ,where N is the number of terms       Â
 Space Complexity : O(1), since no extra space has been taken.
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