# Find N Arithmetic Means between A and B

• Difficulty Level : Medium
• Last Updated : 27 Aug, 2022

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 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 aboveint 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 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 codeA = 20; B = 32; N = 5printAMeans(A, B, N) # This code is contributed by Smitha Dinesh Semwal

## C#

 // C# program to illustrate// n arithmetic mean between // A and Busing 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



## Javascript



Output:

22 24 26 28 30

Time Complexity : O(N) ,where N is the number of terms

Space Complexity : O(1), since no extra space has been taken.

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