Find N Geometric Means between A and B

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Given three integers A, B and N the task is to find N Geometric means between A and B. WE basically need to insert N terms in a Geometric progression. where A and B are first and last terms.
Examples:

Input :  A = 2  B = 32  N = 3
Output : 4 8 16
the geometric progression series as 2,
4, 8, 16 , 32
       
Input : A = 3 B = 81 N = 2
Output : 9 27

Approach :
Let A₁, G₂, G₃, G₄……G_n be N geometric Means between two given numbers A and B . Then A, G₁, G₂ ….. G_n, B will be in Geometric Progression .
So B = (N+2)^th term of the Geometric progression.
Then Here R is the common ratio
B = A*R^N+1
R^N+1 = B/A
R = (B/A)^1/(N+1)
Now we have the value of R
And also we have the value of the first term A
G₁ = AR¹ = A * (B/A)^1/(N+1)
G₂ = AR² = A * (B/A)^2/(N+1)
G₃ = AR³ = A * (B/A)^3/(N+1)
.
.
.
G_N = AR^N = A * (B/A)^N/(N+1)

C++

// C++ program to find n geometric means
// between A and B
#include <bits/stdc++.h>
using namespace std;
 
// Prints N geometric means between
// A and B.
void printGMeans(int A, int B, int N)
{
    // calculate common ratio(R)
    float R = (float)pow(float(B / A), 
                  1.0 / (float)(N + 1));
     
    // for finding N the Geometric 
    // mean between A and B
    for (int i = 1; i <= N; i++) 
        cout << A * pow(R, i) <<" ";    
}
 
// Driver code to test above 
int main()
{
    int A = 3, B = 81, N = 2;
    printGMeans(A, B, N);    
    return 0;
}

Java

// java program to illustrate
// n geometric 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 printGMeans(int A, int B, int N)
    {       
        // Finding the value of R Common ration
        float R = (float)Math.pow((float)(B / A),
                           1.0 / (float)(N + 1));
                            
        // for finding N the Geometric 
        // mean between A and B
        for (int i = 1; i <= N; i++) 
          System.out.print(A * Math.pow(R, i) + " ");
         
    }
 
    // Driver code
    public static void main(String args[])
    {
        int A = 3, B = 81, N = 2;
        printGMeans(A, B, N);
    }
}

Python3

# Python3 program to find
# n geometric means
# between A and B
import math
 
# Prints N geometric means 
# between A and B.
def printGMeans(A, B, N):
     
    # calculate
    # common ratio(R)
    R = (math.pow((B / A),
          1.0 / (N + 1)));
     
    # for finding N the 
    # Geometric mean 
    # between A and B
    for i in range(1, N + 1):
        print(int(A * math.pow(R, i)), 
                           end = " "); 
 
# Driver Code
A = 3;
B = 81;
N = 2;
printGMeans(A, B, N); 
     
# This code is contributed
# by mits

C#

// C# program to illustrate
// n geometric mean between 
// A and B
using System;
 
public class GFG {
 
    // insert function for calculating the means
    static void printGMeans(int A, int B, int N)
    { 
         
        // Finding the value of R Common ration
        float R = (float)Math.Pow((float)(B / A),
                        1.0 / (float)(N + 1));
                             
        // for finding N the Geometric 
        // mean between A and B
        for (int i = 1; i <= N; i++) 
            Console.Write(A * Math.Pow(R, i) + " ");
         
    }
 
    // Driver code
    public static void Main()
    {
        int A = 3, B = 81, N = 2;
         
        printGMeans(A, B, N);
    }
}
 
// This code is contributed by vt_m.

PHP

<?php
// PHP program to find
// n geometric means
// between A and B
 
// Pr$s N geometric means 
// between A and B.
function printGMeans($A, $B, $N)
{
     
    // calculate common ratio(R)
    $R = pow(($B / $A), 
             1.0 / ($N + 1));
     
    // for finding N the Geometric 
    // mean between A and B
    for ($i = 1; $i <= $N; $i++) 
        echo $A * pow($R, $i) ," "; 
}
 
    // Driver Code
    $A = 3;
    $B = 81;
    $N = 2;
    printGMeans($A, $B, $N); 
     
// This code is contributed by anuj_67.
?>

Javascript

<script>
// JavaScript program to illustrate
// n geometric mean between 
// A and B
 
    // insert function for calculating the means
    function printGMeans(A, B, N)
    {       
        // Finding the value of R Common ration
        let R = Math.pow((B / A),
                           1.0 / (N + 1));
                              
        // for finding N the Geometric 
        // mean between A and B
        for (let i = 1; i <= N; i++) 
          document.write(A * Math.pow(R, i) + " ");
           
    }
 
// Driver Code
        let A = 3, B = 81, N = 2;
        printGMeans(A, B, N);
 
// This code is contributed by code_hunt.
</script>

Output :

9 27

Time Complexity : O(N*log(N)) ,in worst case power function takes log(N) time.

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

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