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++ 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 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 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# 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 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
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.