Program for sum of geometric series

A Geometric series is a series with a constant ratio between successive terms. The first term of the series is denoted by a and common ratio is denoted by r. The series looks like this :- a, ar, ar², ar³, ar⁴, . . .. The task is to find the sum of such a series.

Examples :

Input : a = 1
        r = 0.5
        n = 10
Output : 1.99805

Input : a = 2
        r = 2
        n = 15
Output : 65534

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Simple solution to calculate the sum of geometric series.

C++

// A naive solution for calculating sum of 
// geometric series. 
#include<bits/stdc++.h> 
using namespace std; 
  
// function to calculate sum of  
// geometric series 
float sumOfGP(float a, float r, int n) 
{ 
    float sum = 0;  
    for (int i = 0; i < n; i++) 
    { 
        sum = sum + a; 
        a = a * r; 
    } 
    return sum; 
} 
  
// driver function 
int main() 
{ 
    int a = 1; // first term 
    float r = 0.5; // common ratio 
    int n = 10; // number of terms 
  
    cout << sumOfGP(a, r, n) << endl; 
    return 0; 
} 

Java

// A naive solution for calculating sum of 
// geometric series. 
import java.io.*; 
  
class GFG{ 
      
    // function to calculate sum of  
    // geometric series 
    static float sumOfGP(float a, float r, int n) 
    { 
        float sum = 0;  
        for (int i = 0; i < n; i++) 
        { 
            sum = sum + a; 
            a = a * r; 
        } 
        return sum; 
    } 
  
    // driver function 
    public static void main(String args[]) 
    { 
        int a = 1; // first term 
        float r = (float)(1/2.0) ;// common ratio 
        int n = 10 ; // number of terms 
          
        System.out.printf("%.5f",(sumOfGP(a, r, n))); 
    } 
      
} 
  
//This code is contributed by Nikita Tiwari 

Python

# A naive solution for calculating sum of 
# geometric series. 
  
# function to calculate sum of  
# geometric series 
def sumOfGP(a, r, n) : 
      
    sum = 0
    i = 0
    while i < n : 
        sum = sum + a 
        a = a * r 
        i = i + 1
      
    return sum
      
#driver function 
  
a = 1 # first term 
r = (float)(1/2.0) # common ratio 
n = 10 # number of terms 
          
print("%.5f" %sumOfGP(a, r, n)), 
      
# This code is contributed by Nikita Tiwari 

C#

// A naive solution for calculating 
// sum of geometric series. 
using System; 
  
class GFG { 
      
    // function to calculate  
    // sum of geometric series 
    static float sumOfGP(float a,  
                         float r,  
                         int n) 
    { 
        float sum = 0;  
        for (int i = 0; i < n; i++) 
        { 
            sum = sum + a; 
            a = a * r; 
        } 
        return sum; 
    } 
  
    // Driver Code 
    static public void Main () 
    { 
          
        // first term 
        int a = 1;  
          
        // common ratio 
        float r = (float)(1/2.0) ; 
          
        // number of terms 
        int n = 10 ;  
          
        Console.WriteLine((sumOfGP(a, r, n))); 
    } 
      
} 
  
// This code is contributed by Ajit. 

PHP

<?php 
// A naive solution for calculating  
// sum of geometric series. 
  
// function to calculate sum   
// of geometric series 
function sumOfGP($a, $r, $n) 
{ 
    $sum = 0;  
    for ($i = 0; $i < $n; $i++) 
    { 
        $sum = $sum + $a; 
        $a = $a * $r; 
    } 
    return $sum; 
} 
  
// Driver Code 
  
// first term 
$a = 1;  
  
// common ratio 
$r = 0.5;  
  
// number of terms 
$n = 10;  
  
echo(sumOfGP($a, $r, $n)); 
  
// This code is contributed by Ajit. 
?> 

Output :

1.99805

Time Complexity: O(n).

An Efficient solution to solve the sum of geometric series where first term is a and common ration is r
is by the formula :-
sum of series = a(1 – rⁿ)/(1 – r).
Where r = T2/T1 = T3/T2 = T4/T3 . . .
and T1, T2, T3, T4 . . . ,Tn are the first, second, third, . . . ,nth terms respectively.
For example – The series is 2, 4, 8, 16, 32, 64, . . . upto 15 elements. In the above series, find the sum of first 15 elements where
first term a = 2 and common ration r = 4/2 = 2 or = 8/4 = 2
Then,
sum = 2 * (1 – 2¹⁵) / (1 – 2).
sum = 65534

C++

// An Efficient solution to solve sum of  
// geometric series. 
#include<bits/stdc++.h> 
using namespace std; 
  
// function to calculate sum of  
// geometric series 
float sumOfGP(float a, float r, int n) 
{ 
    // calculating and storing sum 
    return (a * (1 - pow(r, n))) / (1 - r); 
} 
  
// driver code 
int main() 
{ 
    float a = 2; // first term 
    float r = 2; // common ratio 
    int n = 15; // number of terms 
  
    cout << sumOfGP(a, r, n); 
    return 0; 
} 

Java

// An Efficient solution to solve sum of  
// geometric series. 
import java.math.*; 
  
class GFG{ 
      
    // function to calculate sum of  
    // geometric series 
    static float sumOfGP(float a, float r, int n) 
    { 
        // calculating and storing sum 
        return (a * (1 - (int)(Math.pow(r, n)))) /  
                                            (1 - r); 
    } 
  
    // driver code 
    public static void main(String args[]) 
    { 
        float a = 2; // first term 
        float r = 2; // common ratio 
        int n = 15; // number of terms 
  
        System.out.println((int)(sumOfGP(a, r, n))); 
    } 
} 
  
//This code is contributed by Nikita Tiwari. 

Python

# An Efficient solution to solve sum of  
# geometric series. 
  
# function to calculate sum of  
# geometric series 
def sumOfGP( a, r, n) : 
      
    # calculating and storing sum 
    return (a * (1 - pow(r, n))) / (1 - r) 
      
      
# driver code 
a = 2 # first term 
r = 2 # common ratio 
n = 15 # number of terms 
  
print sumOfGP(a, r, n) 
  
# This code is contributed by Nikita Tiwari. 

C#

// C# program to An Efficient solution 
// to solve sum of geometric series. 
using System; 
  
class GFG { 
      
    // function to calculate sum of  
    // geometric series 
    static float sumOfGP(float a, float r, int n) 
    { 
        // calculating and storing sum 
        return (a * (1 - (int)(Math.Pow(r, n)))) /  
                                       (1 - r); 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        float a = 2; // first term 
        float r = 2; // common ratio 
        int n = 15; // number of terms 
  
        Console.Write((int)(sumOfGP(a, r, n))); 
    } 
} 
  
// This code is contributed by Nitin Mittal. 

PHP

<?php 
// An Efficient solution to solve  
// sum of geometric series. 
  
// function to calculate sum  
// of geometric series 
function sumOfGP($a, $r, $n) 
{ 
    // calculating and storing sum 
    return ($a * (1 - pow($r, $n))) /  
                         (1 - $r); 
} 
  
// Driver Code 
  
// first term 
$a = 2;  
  
// common ratio 
$r = 2;  
  
// number of terms 
$n = 15;  
  
echo(sumOfGP($a, $r, $n)); 
  
// This code is contributed by Ajit. 
?> 

Output :

Time Complexity: Depends on implementation of pow() function in C/C++. In general, we can compute integer powers in O(Log n) time.

This article is contributed by Dharmendra kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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