# 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, ar ^{2}, ar^{3}, ar^{4}, . . .**. 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++

// 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

// 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

# 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#

// 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 – r^{n})/(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 – 2^{15}) / (1 – 2).

sum = 65534

## C++

// 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

// 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

# 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#

// 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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