Given a number N, the first term ‘a’ and a common ratio ‘r’ of a Geometric Progression, the task is to find the product of first N terms of the given Geometric Progression.
Examples:
Input: a = 1, r = 2, N = 4
Output: 64
Explanation:
First four term for the above G.P. is 1, 2, 4, 8.
Product of four numbers is 1*2*4*8 = 64Input: a = 1, r = 0.5, N = 3
Output: 0.125
Explanation:
First three term for the above G.P. is 1, 1/2, 1/4.
Product of four numbers is 1*(1/2)*(1/4) = 1/8 = 0.125
Naive Approach: The idea is to find all the N terms of the given geometric progression and multiply all the terms obtained.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate product of// geometric seriesfloat productOfGP(float a, float r, int n){ // Initialise final product with 1 float product = 1; for (int i = 0; i < n; i++) { // Multiply product with each // term stored in a product = product * a; a = a * r; } // Return the final product return product;}// Driver Codeint main(){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call cout << productOfGP(a, r, N);} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to calculate product of// geometric seriesstatic float productOfGP(float a, float r, int n){ // Initialise final product with 1 float product = 1; for (int i = 0; i < n; i++) { // Multiply product with each // term stored in a product = product * a; a = a * r; } // Return the final product return product;}// Driver Codepublic static void main(String args[]){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call System.out.print(productOfGP(a, r, N));}}// This code is contributed by Code_Mech |
Python3
# Python3 program for the above approach# Function to calculate product of# geometric seriesdef productOfGP(a, r, n): # Initialise final product with 1 product = 1; for i in range(0, n): # Multiply product with each # term stored in a product = product * a; a = a * r; # Return the final product return product; # Driver code# Given first term# and common ratioa = 1r = 2;# Number of termsN = 4;# Function Callprint(productOfGP(a, r, N))# This code is contributed by Pratima Pandey. |
C#
// C# program for the above approachusing System;class GFG{// Function to calculate product of// geometric seriesstatic float productOfGP(float a, float r, int n){ // Initialise final product with 1 float product = 1; for (int i = 0; i < n; i++) { // Multiply product with each // term stored in a product = product * a; a = a * r; } // Return the final product return product;}// Driver Codepublic static void Main(){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call Console.Write(productOfGP(a, r, N));}}// This code is contributed by Code_Mech |
Javascript
<script> // Javascript program for the above approach // Function to calculate product of // geometric series function productOfGP(a, r, n) { // Initialise final product with 1 let product = 1; for (let i = 0; i < n; i++) { // Multiply product with each // term stored in a product = product * a; a = a * r; } // Return the final product return product; } // Given first term // and common ratio let a = 1, r = 2; // Number of terms let N = 4; // Function Call document.write(productOfGP(a, r, N)); // This code is contributed by divyesh072019.</script> |
Output:
64
Time Complexity: O(N), where N is the given integer.
Auxiliary Space: O(1), as constant extra space is required.
Efficient Approach: The efficient approach is to observe that the product of all the N terms of the given geometric progression forms a formula:
The general G.P. is
Product upto 1st term =
Product upto 2nd term =
Product upto 3rd term =
Product upto 4th term =
.
.
.
Product upto N terms =
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate product of// geometric seriesfloat productOfGP(float a, float r, int n){ // Return the final product with the // above formula return pow(a, n) * pow(r, n * (n - 1) / 2);}// Driver Codeint main(){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call cout << productOfGP(a, r, N);} |
Java
// Java program for the above approachclass GFG{ // Function to calculate product of// geometric seriesstatic float productOfGP(float a, float r, int n){ // Return the final product with the // above formula return (float)Math.pow(a, n) * (float)Math.pow(r, n * (n - 1) / 2);}// Driver Codepublic static void main(String s[]){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call System.out.println(productOfGP(a, r, N));}}// This code is contributed by rutvik_56 |
Python3
# Python3 program for the above approach# Function to calculate product of# geometric seriesdef productOfGP(a, r, n): # Return the final product with the # above formula return pow(a, n) * pow(r, n * (n - 1) // 2);# Driver Code# Given first term# and common ratioa = 1; r = 2;# Number of termsN = 4;# Function Callprint(productOfGP(a, r, N));# This code is contributed by Code_Mech |
C#
// C# program for the above approachusing System;class GFG{ // Function to calculate product of// geometric seriesstatic float productOfGP(float a, float r, int n){ // Return the final product with the // above formula return (float)Math.Pow(a, n) * (float)Math.Pow(r, n * (n - 1) / 2);}// Driver Codepublic static void Main(String[] args){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call Console.WriteLine(productOfGP(a, r, N));}}// This code is contributed by shivanisinghss2110 |
Javascript
<script>// Javascript program for the above approach// Function to calculate product of// geometric seriesfunction productOfGP(a, r, n){ // Return the final product with the // above formula return Math.pow(a, n) * Math.pow(r, n * (n - 1) / 2);}// Driver code// Given first term// and common ratiolet a = 1, r = 2;// Number of termslet N = 4;// Function Calldocument.write(productOfGP(a, r, N));// This code is contributed by mukesh07 </script> |
Output:
64
Time Complexity: O(logN), for using pow function.
Auxiliary space: O(1), as constant extra space is required.
Another Approach:
Let there be a GP of N terms as followed:
Product of 1st and last term =
=
Product of 2nd and 2nd last term =
=
Product of 3rd and 3rd last term =
=
and so on…
In a geometric progression, the product of the first and the last term is the same as the product of the second and second last term and so on, no matter how many terms we are considering in a geometric progression.
Therefore, ifand
is the first and last terms of a GP, then the product of first N terms of the GP will be
and 
is the first and last terms of a GP, then the product of first N terms of the GP will be 
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate product of N// terms of geometric seriesfloat productOfGP(float a, float r, int n){ // Find the product of first // and the last term int an = a * pow(r, n - 1); // Return the sqrt of the above // expression to find the product return sqrt(pow(a * an, n));}// Driver Codeint main(){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call cout << productOfGP(a, r, N);} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to calculate product of N// terms of geometric seriesstatic float productOfGP(float a, float r, int n){ // Find the product of first // and the last term int an = (int)(a * (int)(Math.pow(r, n - 1))); // Return the sqrt of the above // expression to find the product return (int)Math.sqrt((int)Math.pow(a * an, n));}// Driver Codepublic static void main(String args[]){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call System.out.print(productOfGP(a, r, N));}}// This code is contributed by Code_Mech |
Python3
# Python3 program for the above approach import math# Function to calculate product of N# terms of geometric seriesdef productOfGP(a, r, n): # Find the product of first # and the last term an = a * pow(r, n - 1); # Return the sqrt of the above # expression to find the product return (math.sqrt(pow(a * an, n))) # Driver code# Given first term# and common ratioa = 1r = 2;# Number of termsN = 4;# Function Callprint(productOfGP(a, r, N))# This code is contributed by Pratima Pandey. |
C#
// C# program for the above approachusing System;class GFG{// Function to calculate product of N// terms of geometric seriesstatic float productOfGP(float a, float r, int n){ // Find the product of first // and the last term int an = (int)(a * (int)(Math.Pow(r, n - 1))); // Return the sqrt of the above // expression to find the product return (int)Math.Sqrt((int)Math.Pow(a * an, n));}// Driver Codepublic static void Main(){ // Given first term // and common ratio float a = 1, r = 2; // Number of terms int N = 4; // Function Call Console.Write(productOfGP(a, r, N));}}// This code is contributed by Code_Mech |
Javascript
<script>// Javascript program for the above approach// Function to calculate product of N// terms of geometric seriesfunction productOfGP(a, r, n){ // Find the product of first // and the last term let an = a * Math.pow(r, n - 1); // Return the sqrt of the above // expression to find the product return Math.sqrt(Math.pow(a * an, n));}// Driver code// Given first term// and common ratiolet a = 1, r = 2;// Number of termslet N = 4;// Function Calldocument.write(productOfGP(a, r, N));// This code is contributed by divyeshrabadiya07</script> |
Output:
64
Time Complexity: O(logN), for using pow and sqrt functions.
Auxiliary space: O(1), as constant extra space is required.
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