Find N Geometric Means between A and B - GeeksforGeeks

07 Apr, 2021
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......Gn be N geometric Means between two given numbers A and B . Then A, G1, G2 ..... Gn, 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*RN+1 RN+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) . . . GN = ARN = A * (B/A)N/(N+1) 
C++
Java
Python3
C#
PHP
Javascript
// 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 aboveint main(){    int A = 3, B = 81, N = 2;    printGMeans(A, B, N);       return 0;}
// java program to illustrate// n geometric mean between// A and Bimport 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 program to find# n geometric means# between A and Bimport 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 CodeA = 3;B = 81;N = 2;printGMeans(A, B, N);     # This code is contributed# by mits
// C# program to illustrate// n geometric mean between// A and Busing 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 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.?>
<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 
 
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Mithun Kumar
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