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++ 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 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 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# 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 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. ?> |
<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 code let 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.