在A和B之间找到N个几何均值

给定三个整数A，B和N，任务是在A和B之间找到N个几何均值。我们基本上需要在几何级数中插入N个项。其中A和B是第一项和最后一项。
例子：

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

方法 ：
设A ₁ ，G ₂ ，G ₃ ，G ₄ ……G _n为两个给定数字A和B之间的N个几何均值。然后，A，G ₁ ，G ₂ …..G _n ，B将处于几何_级数。
因此，B =几何级数的^第(N + 2)^个项。
那么这里R是公比
B = A * R ^{N + 1}
R ^{N + 1} = B / A
R =(B / A) ^{1 /(N + 1)}
现在我们有了R的值
还有第一项A的值
G ₁ = AR ¹ = A *(B / A) ^{1 /(N + 1)}
G ₂ = AR ² = A *(B / A) ^{2 /(N + 1)}
G ₃ = AR ³ = 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 
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.

Javascript

// 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.

输出 ：

9 27