Harmonic Progression
A sequence of numbers is called a Harmonic progression if the reciprocal of the terms are in AP. In simple terms, a, b, c, d, e, f are in HP if 1/a, 1/b, 1/c, 1/d, 1/e, 1/f are in AP. For example, 1/a, 1/(a+d), 1/(a+2d), and so on are in HP because a, a + d, a + 2d are in AP.
Fact about Harmonic Progression :
- In order to solve a problem on Harmonic Progression, one should make the corresponding AP series and then solve the problem.
- As the nth term of an A.P is given by an = a + (n-1)d, So the nth term of an H.P is given by 1/ [a + (n -1) d].
- For two numbers, if A, G and H are respectively the arithmetic, geometric and harmonic means, then
- A ≥ G ≥ H
- A H = G2, i.e., A, G, H are in GP
- If we need to find three numbers in a H.P. then they should be assumed as 1/a–d, 1/a, 1/a+d
- Majority of the questions of H.P. are solved by first converting them into A.P
Formula of Harmonic Progression:
How we check whether a series is harmonic progression or not?
The idea is to reciprocal the given array or series. After reciprocal, check if differences between consecutive elements are same or not. If all differences are same, Arithmetic Progression is possible. So as we know if the reciprocal of the terms are in AP then given a sequence of series is in H.P. Let’s take a series 1/5, 1/10, 1/15, 1/20, 1/25 and check whether it is a harmonic progression or not. Below is the implementation:
C++
// CPP program to check if a given // array can form harmonic progression #include<bits/stdc++.h> using namespace std; bool checkIsHP(vector< double > &arr) { int n = arr.size(); if (n == 1) { return true ; } // Find reciprocal of arr[] vector< int > rec; for ( int i = 0; i < n; i++) { rec.push_back((1 / arr[i])); } // return (rec); // After finding reciprocal, check if // the reciprocal is in A. P. // To check for A.P., first Sort the // reciprocal array, then check difference // between consecutive elements sort(rec.begin(), rec.end()); int d = (rec[1]) - (rec[0]); for ( int i = 2; i < n; i++) { if (rec[i] - rec[i - 1] != d) { return false ; } } return true ; } // Driver Code int main() { // series to check whether it is in H.P vector< double > arr = {1 / 5, 1 / 10, 1 / 15, 1 / 20, 1 / 25}; // Checking a series is in H.P or not if (checkIsHP(arr)) { cout << "Yes" << std::endl; } else { cout << "No" <<endl; } return 0; } // This code is contributed by mits |
Java
// Java program to check if a given // array can form harmonic progression import java.util.*; class GFG { static boolean checkIsHP( double []arr) { int n = arr.length; if (n == 1 ) return true ; // Find reciprocal of arr[] ArrayList<Integer> rec = new ArrayList<Integer>(); for ( int i = 0 ; i < n; i++) rec.add(( int )( 1 / arr[i])); // return (rec); // After finding reciprocal, check if // the reciprocal is in A. P. // To check for A.P., first Sort the // reciprocal array, then check difference // between consecutive elements Collections.sort(rec); int d = ( int )rec.get( 1 ) - ( int )rec.get( 0 ); for ( int i = 2 ; i < n; i++) if (rec.get(i) - rec.get(i - 1 ) != d) return false ; return true ; } // Driver code public static void main(String[] args) { // series to check whether it is in H.P double arr[] = { 1 / 5 , 1 / 10 , 1 / 15 , 1 / 20 , 1 / 25 }; // Checking a series is in H.P or not if (checkIsHP(arr)) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by mits |
Python3
# Python3 program to check if a given # array can form harmonic progression def checkIsHP(arr): n = len (arr) if (n = = 1 ): return True # Find reciprocal of arr[] rec = [] for i in range ( 0 , len (arr)): a = 1 / arr[i] rec.append(a) return (rec) # After finding reciprocal, check if the # reciprocal is in A. P. # To check for A.P., first Sort the # reciprocal array, then check difference # between consecutive elements rec.sort() d = rec[ 1 ] - rec[ 0 ] for i in range ( 2 , n): if (rec[i] - rec[i - 1 ] ! = d): return False return True # Driver code if __name__ = = '__main__' : # series to check whether it is in H.P arr = [ 1 / 5 , 1 / 10 , 1 / 15 , 1 / 20 , 1 / 25 ] # Checking a series is in H.P or not if (checkIsHP(arr)): print ( "Yes" ) else : print ( "No" ) |
C#
// C# program to check if a given // array can form harmonic progression using System; using System.Collections; class GFG { static bool checkIsHP( double [] arr) { int n = arr.Length; if (n == 1) return true ; // Find reciprocal of arr[] ArrayList rec = new ArrayList(); for ( int i = 0; i < n; i++) rec.Add(( int )(1 / arr[i])); // return (rec); // After finding reciprocal, check if // the reciprocal is in A. P. // To check for A.P., first Sort the // reciprocal array, then check difference // between consecutive elements rec.Sort(); int d = ( int )rec[1] - ( int )rec[0]; for ( int i = 2; i < n; i++) if (( int )rec[i] - ( int )rec[i - 1] != d) return false ; return true ; } // Driver code public static void Main() { // series to check whether it is in H.P double [] arr = { 1/5, 1/10, 1/15, 1/20, 1/25 }; // Checking a series is in H.P or not if (checkIsHP(arr)) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by mits |
PHP
<?php // PHP program to check if a given // array can form harmonic progression function checkIsHP( $arr ) { $n = count ( $arr ); if ( $n == 1) return true; // Find reciprocal of arr[] $rec = array (); for ( $i =0; $i < count ( $arr ); $i ++) { $a = 1 / $arr [ $i ]; array_push ( $rec , $a ); } return ( $rec ); // After finding reciprocal, check if the // reciprocal is in A. P. // To check for A.P., first Sort the // reciprocal array, then check difference // between consecutive elements sort( $rec ); $d = $rec [1] - $rec [0]; for ( $i =2; $i < $n ; $i ++) if ( $rec [ $i ] - $rec [ $i -1] != $d ) return false; return true; } // Driver code // series to check whether it is in H.P $arr = array ( 1/5, 1/10, 1/15, 1/20, 1/25 ); // Checking a series is in H.P or not if (checkIsHP( $arr )) print ( "Yes" ); else print ( "No" ); // This code is contributed by mits ?> |
Javascript
<script> // JavaScript program to check if a given // array can form harmonic progression function checkIsHP(arr) { let n = arr.length; if (n == 1) { return true ; } // Find reciprocal of arr[] let rec = []; for (let i = 0; i < n; i++) { rec.push((1 / arr[i])); } // return (rec); // After finding reciprocal, check if // the reciprocal is in A. P. // To check for A.P., first Sort the // reciprocal array, then check difference // between consecutive elements rec.sort((a,b) => a - b); let d = (rec[1]) - (rec[0]); for (let i = 2; i < n; i++) { if (rec[i] - rec[i - 1] != d) { return false ; } } return true ; } // Driver Code // series to check whether it is in H.P let arr = [1 / 5, 1 / 10, 1 / 15, 1 / 20, 1 / 25]; // Checking a series is in H.P or not if (checkIsHP(arr)) { document.write( "Yes" ); } else { document.write( "No" ); } </script> |
Output:
Yes
Time Complexity: O(n Log n).
Auxiliary Space: O(n)
Basic Program related to Harmonic Progression
- Harmonic progression Sum
- Program to find the Nth Harmonic Number
- Sum of series (n/1) + (n/2) + (n/3) + (n/4) +…….+ (n/n)
- Check if the given number is Ore number or not
- Expected Number of Trials until Success
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