Sum of Geometric Progression series (original) (raw)

Last Updated : 22 May, 2026

Given **n, **a and **r as the number of terms, first term and common ratio respectively of a Geometric Progression series. Find the sum of the series up to **nth term.

The Geometric Progression series looks like :- a, ar, ar2, ar3, ar4, .....

**Examples :

**Input: n = 3, a = 3, r = 2
**Output: 21
**Explanation: Series up to 3rd term is 3 6 12, so sum will be 21.

**Input : n=3, a = 1, r = 2
**Output : 7
**Explanation: Series up to 3rd term is 1 2 4, so sum will be 7.

Try It Yourself

Table of Content

[Naive Approach] Iterative Summation Method - O(N) Time and O(1) Space
[Expected Approach] Direct Formula Method - O(log(n)) Time and O(1) Space

[Naive Approach] Iterative Summation Method - O(N) Time and O(1) Space

A simple solution is to one by one add terms to calculate the sum of geometric series.

Consider we have n = 5, a = 2 and r = 2.

Initializing sum = 0, and then iterating through all terms

i = 0: sum becomes 0 + 2 = 2, then a becomes 2 * 2 = 4
i = 1: sum becomes 2 + 4 = 6, then a becomes 4 * 2 = 8.
i = 2: sum becomes 6 + 8 = 14, then a becomes 8 * 2 = 16.
i = 3: sum becomes 14 + 16 = 30, then a becomes 16 * 2 = 32.
i = 4: sum becomes 30 + 32 = 62, then a becomes 32 * 2 = 64. C++ `

#include <bits/stdc++.h> using namespace std;

// function to calculate sum of // geometric series int sumOfGP(int n, int a, int r) { int sum = 0;

// iterate n times to generate and 
// add each term in the progression
for (int i = 0; i < n; i++)
{
    sum = sum + a;

    // multiplying with r to get next term
    a = a * r;
}
return sum;

}

// driver function int main() { int a = 3;
int r = 2; int n = 3;

cout << sumOfGP(n, a, r) << endl;
return 0;

}

Java

import java.io.*;

class GFG{

// function to calculate sum of 
// geometric series
static int sumOfGP(int n, int a, int r)
{
    int sum = 0; 
    
    // iterate n times to generate and 
    // add each term in the progression
    for (int i = 0; i < n; i++)
    {
        sum = sum + a;
        
        // multiplying with r to get next term
        a = a * r;
    }
    return sum;
}

// driver function
public static void main(String args[])
{
    int a = 3; 
    int r = 2; 
    int n = 3 ; 
    
    System.out.printf(sumOfGP(n, a, r));
}

}

Python

function to calculate sum of

geometric series

def sumOfGP(n, a, r) :

sum = 0
i = 0

# iterate n times to generate and
# add each term in the progression
while i < n :
    sum = sum + a
    
    # multiplying with r to get next term
    a = a * r
    i = i + 1

return sum

#driver function

a = 3 r = 2 n = 3

print(sumOfGP(n, a, r)),

C#

using System;

class GFG {

// function to calculate 
// sum of geometric series
static int sumOfGP(int n, int a, int r)
{
    int sum = 0; 
    
    // iterate n times to generate and 
    // add each term in the progression
    for (int i = 0; i < n; i++)
    {
        sum = sum + a;
        
        // multiplying with r to get next term
        a = a * r;
    }
    return sum;
}

// Driver Code
static public void Main ()
{
    
    int a = 3; 
    int r = 2;
    int n = 3; 
    
    Console.WriteLine((sumOfGP(n, a, r)));
}

}

JavaScript

// function to calculate sum // of geometric series function sumOfGP(n, a, r) { let sum = 0;

// iterate n times to generate and 
// add each term in the progression
for (let i = 0; i < n; i++) {
    sum = sum + a;
    
    // multiplying with r to get next term
    a = a * r;
}
return sum;

}

// Driver code let a = 3;
let r = 2; let n = 3;

console.log(sumOfGP(n, a, r))

[Expected Approach] Direct Formula Method - O(log(n)) Time and O(1) Space

An efficient solution to solve the sum of geometric series where first term is a and common ration isr is by the formula :- sum of series = a(1 - rn)/(1 - r). Where r = T2/T1 = T3/T2 = T4/T3 . . . Here T1, T2, T3, T4 . . . ,Tn are the first, second, third, . . . ,nth terms respectively.

sumofthenthtermofGP

Consider we have n = 5, a = 2 and r = 2.

Series is 2, 4, 8, 16, 32 which sum up to 62 = 2 * (25 - 1) / (2 - 1).

C++ `

#include <bits/stdc++.h> using namespace std;

// function to calculate sum of // geometric series int sumOfGP(int n, int a, int r) {

// if common ratio is 1, the series 
// is simply a, a, a, ...(n times)
if (r == 1)
{
    return a * n;
}

// applying standard geometric series 
// formula : a * (r^n - 1) / (r - 1)
return (a * (pow(r, n) - 1)) / (r - 1);

}

// driver code int main() { int a = 3;
int r = 2;
int n = 3;

cout << sumOfGP(n, a, r);
return 0;

}

Java

import java.math.*;

class GFG {

// function to calculate sum of
// geometric series
static int sumOfGP(int n, int a, int r)
{
    
    // if common ratio is 1, the series 
    // is simply a, a, a, ...(n times)
    if (r == 1) {
        return a * n;
    }

    // applying standard geometric series 
    // formula : a * (r^n - 1) / (r - 1)
    return (a * (int)(1 - (Math.pow(r, n)))) / (1 - r);
}

// driver code
public static void main(String args[])
{
    int a = 3;  
    int r = 2; 
    int n = 3; 

    System.out.println((sumOfGP(n, a, r)));
}

}

Python

function to calculate sum of

geometric series

def sumOfGP(n, a, r) :

# if common ratio is 1, the series 
# is simply a, a, a, ...(n times)
if r == 1:
    return a * n

# applying standard geometric series 
# formula : a * (r^n - 1) / (r - 1) 
return (a * (pow(r, n) - 1)) // (r - 1)

driver code

a = 3
r = 2
n = 3

print sumOfGP(n, a, r)

C#

using System;

class GFG {

// function to calculate sum of 
// geometric series
static int sumOfGP(int n, int a, int r)
{
    
    // if common ratio is 1, the series 
    // is simply a, a, a, ...(n times)
    if (r == 1) {
        return a * n;
    }

    // applying standard geometric series 
    // formula : a * (r^n - 1) / (r - 1)
    return (a * (1 - (Math.Pow(r, n)))) / (1 - r);
}

// Driver Code
public static void Main()
{
    int a = 3;  
    int r = 2;  
    int n = 3;  

    Console.Write(sumOfGP(n, a, r));
}

}

JavaScript

// function to calculate sum of // geometric series function sumOfGP(n, a, r) {

// if common ratio is 1, the series is
// simply a, a, a,...(n times)
if (r == 1) {
    return a * n;
}

// applying standard geometric series
// formula : a * (r^n - 1) / (r - 1)
return (a * (Math.pow(r, n)) - 1) / (r - 1);

}

// Driver code function main() { let a = 3; let r = 2; let n = 3;

console.log(sumOfGP(n, a, r));

}

// Run the main function main();