Find N Arithmetic Means between A and B (original) (raw)

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 <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

B,B, B,N) { // calculate common // difference(d) d=(d = (d=(B - A)/(A) / (A)/(N + 1); // for finding N the arithmetic // mean between A and B for ($i = 1; i<=i <= i<=N; $i++) echo ($A + i∗i * i∗d) ," "; } // Driver Code A=20;A = 20; A=20;B = 32; $N = 5; printAMeans($A, B,B, B,N); // This code is Contributed by vt_m. ?>

JavaScript

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

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