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Program for sum of geometric series
  • Difficulty Level : Basic
  • Last Updated : 27 Apr, 2018
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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

A Simple solution to calculate the sum of geometric series.

C++




// A naive solution for calculating sum of
// geometric series.
#include
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





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
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





Output :
65534

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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