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 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 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.?> |
Javascript
<script>// JavaScript 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) 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)," ") }// Driver codelet A = 20, B = 32, N = 5;printAMeans(A, B, N) // This code is contributed by Shinjanpatra</script> |
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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