# Find N Arithmetic Means between A and B

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 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



## 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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