Program for sum of geometric series
Last Updated :
13 Feb, 2025
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 = 3
Output : 1.75
We get sum as 1 + 1 * 0.5 + 1 * 0.5 * 0.5 = 1 + 0.5 + 0.25 = 1.75
Input : a = 2
r = 2
n = 15
Output : 65534
[Naive Approach] Iterative Summation Method
A Simple solution is to one by one add terms 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 = 3; // number of terms
cout << sumOfGP(a, r, n) << endl;
return 0;
}
// A naive solution for calculating sum of
// geometric series.
#include <stdio.h>
// 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
printf("%f\n", sumOfGP(a, r, n));
return 0;
}
// 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)));
}
}
# 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)),
// 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)));
}
}
// function to calculate sum of
// geometric series
function sumOfGP(a, r, n) {
let sum = 0;
for (let i = 0; i < n; i++) {
sum = sum + a;
a = a * r;
}
return sum;
}
// Driver code
let a = 1; // first term
let r = 0.5; // common ratio
let n = 10; // number of terms
console.log(sumOfGP(a, r, n))
<?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));
?>
Output :
1.99805
Time Complexity : O(n)
Auxiliary Space : O(1)
[Expected Approach] Direct Formula Method
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 . . . Here 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;
}
// An Efficient solution to solve sum of
// geometric series.
#include <stdio.h>
#include <math.h>
// 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
printf("%f", sumOfGP(a, r, n));
return 0;
}
// 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)));
}
}
# 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)
// 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)));
}
}
// Function to calculate sum of geometric series
function sumOfGP(a, r, n) {
// calculating and storing sum
return (a * (1 - Math.pow(r, n))) / (1 - r);
}
// Driver code
function main() {
let a = 2; // first term
let r = 2; // common ratio
let n = 15; // number of terms
console.log(sumOfGP(a, r, n));
}
// Run the main function
main();
<?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));
?>
Output :
65534
Time Complexity: O(Log n) (Depends on implementation of pow() function in C/C++. In general, we can compute integer powers in log(n) time.
Auxiliary Space : O(1)
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