Program to print GP (Geometric Progression)

Given first term (a), common ratio (r) and a integer n of the Geometric Progression series, the task is to print the n terms of the series.
Examples:

Input : a = 2 r = 2, n = 4
Output : 2 4 8 16

Approach :

We know the Geometric Progression series is like = 2, 4, 8, 16, 32 ……. 
In this series 2 is the starting term of the series . 
Common ratio = 4 / 2 = 2 (ratio common in the series). 
so we can write the series as :
t1 = a1 
t2 = a1 * r^(2-1)
t3 = a1 * r^(3-1)
t4 = a1 * r^(4-1)
. 
. 
. 
. 
tN = a1 * r^(n-1)

To print the Geometric Progression series we use the simple formula . 

TN = a1 * r(n-1)

CPP

// CPP program to print GP.
#include <bits/stdc++.h>
using namespace std;

void printGP(int a, int r, int n)
{
    int curr_term;
    for (int i = 0; i < n; i++) {
        curr_term = a * pow(r, i);
        cout << curr_term << " ";
    }
}

// Driver code
int main()
{
    int a = 2; // starting number
    int r = 3; // Common ratio
    int n = 5; // N th term to be find
    printGP(a, r, n);
    return 0;
}

Java

// Java program to print GP.
class GFG {
    static void printGP(int a, int r, int n)
    {
        int curr_term;
        for (int i = 0; i < n; i++) {
            curr_term = a * (int)Math.pow(r, i);
            System.out.print(curr_term + " ");
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int a = 2; // starting number
        int r = 3; // Common ratio
        int n = 5; // N th term to be find
        printGP(a, r, n);
    }
}
// This code is contributed by Anant Agarwal.

Python3

# Python 3 program to print GP.

def printGP(a, r, n):
    for i in range(0, n):
        curr_term = a * pow(r, i)
        print(curr_term, end =" ")


# Driver code

a = 2 # starting number
r = 3 # Common ratio
n = 5 # N th term to be find

printGP(a, r, n)

# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to print GP.
using System;

class GFG {

    static void printGP(int a, int r, int n)
    {

        int curr_term;

        for (int i = 0; i < n; i++) {
            curr_term = a * (int)Math.Pow(r, i);
            Console.Write(curr_term + " ");
        }
    }

    // Driver code
    public static void Main()
    {

        int a = 2; // starting number
        int r = 3; // Common ratio
        int n = 5; // N th term to be find

        printGP(a, r, n);
    }
}

// This code is contributed by vt_m.

Javascript

<script>

// JavaScript program to print GP.

function printGP(a, r, n)
{
    let curr_term;
    for (let i = 0; i < n; i++) {
        curr_term = a * Math.pow(r, i);
        document.write(curr_term + " ");
    }
}

// Driver code

    let a = 2; // starting number
    let r = 3; // Common ratio
    let n = 5; // N th term to be find
    printGP(a, r, n);


// This code is contributed by Surbhi Tyagi

</script>

PHP

<?php
// PHP program to print GP.

// function to print GP
function printGP($a, $r, $n)
{
    for ($i = 0; $i < $n; $i++)
    {
        $curr_term = $a * pow($r, $i);
        echo $curr_term, " ";
    }
}

    // Driver Code

    // starting number
    $a = 2;

    // Common ratio
    $r = 3;

    // N th term to be find
    $n = 5;
    printGP($a, $r, $n);

// This code is contributed by ajit.
?>

Output

2 6 18 54 162 

Time Complexity: O(nlog₂n), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Approach 2: Using recursion to calculate each term of the GP and printing each term.

 The printGP(int a, int r, int n) function takes three integer inputs a, r, and n, and recursively prints the first n terms of a geometric progression with first term a and common ratio r. If n is 0, the function returns without printing anything. Otherwise, it first calculates the current term currTerm using the formula currTerm = a * pow(r, n – 1), and then calls itself recursively with n-1 as the new input argument. Finally, it prints the current term on the console.

•    In the main() function, we initialize the first term a to 2, common ratio r to 3, and the number of terms to print n to 5.
•    We use cout to print the message “GP Series: ” on the console.
•    We call the printGP() function with a, r, and n as input arguments to recursively print the first n terms of the geometric progression.
•    We use cout to insert a newline character after the output.
•    Finally, we return 0 to indicate successful completion of the program.
• Overall, this code prints the first n terms of a geometric progression with given first term a, common ratio r, and number of terms to print n using a recursive function, and prints the result on the console.

C++

#include <iostream>
#include <cmath>
using namespace std;

void printGP(int a, int r, int n)
{
    if (n == 0)
    {
        return;
    }
    int currTerm = a * pow(r, n - 1);
    printGP(a, r, n - 1);
    cout << currTerm << " ";
}

int main()
{
    int a = 2;  // first term
    int r = 3;  // common ratio
    int n = 5;  // number of terms
    cout << "GP Series: ";
    printGP(a, r, n);
    cout << endl;
    return 0;
}

Java

import java.lang.Math;

public class Main {
    public static void printGP(int a, int r, int n) {
        if (n == 0) {
            return;
        }
        int currTerm = a * (int)Math.pow(r, n - 1);
        printGP(a, r, n - 1);
        System.out.print(currTerm + " ");
    }

    public static void main(String[] args) {
        int a = 2;  // first term
        int r = 3;  // common ratio
        int n = 5;  // number of terms
        System.out.print("GP Series: ");
        printGP(a, r, n);
        System.out.println();
    }
}

Python3

import math

def printGP(a, r, n):
    if n == 0:
        return
    currTerm = a * pow(r, n - 1)
    printGP(a, r, n - 1)
    print(currTerm, end=" ")

a = 2  # first term
r = 3  # common ratio
n = 5  # number of terms
print("GP Series:", end=" ")
printGP(a, r, n)
print()

C#

using System;

public class Program
{
    public static void Main()
    {
        int a = 2;  // first term
        int r = 3;  // common ratio
        int n = 5;  // number of terms
        Console.Write("GP Series: ");
        PrintGP(a, r, n);
        Console.WriteLine();
    }

    static void PrintGP(int a, int r, int n)
    {
        if (n == 0)
        {
            return;
        }
        int currTerm = a * (int)Math.Pow(r, n - 1);
        PrintGP(a, r, n - 1);
        Console.Write(currTerm + " ");
    }
}

Javascript

function printGP(a, r, n) {
  // If n is 0, return nothing
  if (n === 0) {
    return;
  }

  // Calculate the current term
  const currTerm = a * Math.pow(r, n - 1);

  // Recursively call printGP with n - 1 until n is 0
  printGP(a, r, n - 1);

  // Print the current term
  console.log(currTerm + " ");
}

const a = 2; // first term
const r = 3; // common ratio
const n = 5; // number of terms

console.log("GP Series: ");
printGP(a, r, n);
console.log("");

Output

GP Series: 2 6 18 54 162 

Time complexity :- O(N)
Space complexity :- O(N)

Approach 3(Using Loop): To solve the problem follow the below idea:

• Initialize a counter variable i to 0.
• Use a while loop to iterate over the first n terms of the series, printing out each term and multiplying the previous term by the common ratio to get the next term.

Below is the implementation of the above approach:

C++

#include <iostream>
using namespace std;

void printGP(int a, int r, int n) {
    for (int i = 0; i < n; i++) {
        cout << a << " ";
        a *= r;
    }
}

int main() {
    // Starting number
    int a = 2;

    // Common ratio
    int r = 3;

    // Nth term to be found
    int n = 5;

    cout << "GP Series: ";
    // Function call
    printGP(a, r, n);

    return 0;
}

Java

// java program to find nth term
// of geometric progression
import java.io.*;
import java.lang.*;

class GFG {

    public static void printGP(int a, int r, int n)
    {
        int i = 0;
        while (i < n) {
            System.out.print(a + " ");
            a *= r;
            i++;
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        // starting number
        int a = 2;

        // Common ratio
        int r = 3;

        // N th term to be find
        int n = 5;

        System.out.print("GP Series: ");
        // Function call
        printGP(a, r, n);

    }
}

Python

# Function to print geometric progression (GP) series
def print_gp(a, r, n):
    for i in range(n):
        print(a),
        a *= r

# Main function


def main():
    # Starting number
    a = 2

    # Common ratio
    r = 3

    # Nth term to be found
    n = 5

    print("GP Series: "),

    # Function call
    print_gp(a, r, n)


# Run the main function
if __name__ == "__main__":
    main()

C#

using System;

class Program
{
    static void PrintGP(int a, int r, int n)
    {
        for (int i = 0; i < n; i++)
        {
            Console.Write(a + " ");
            a *= r;
        }
    }

    static void Main()
    {
        // Starting number
        int a = 2;

        // Common ratio
        int r = 3;

        // Nth term to be found
        int n = 5;

        Console.Write("GP Series: ");
        // Function call
        PrintGP(a, r, n);
    }
}

Javascript

function printGP(a, r, n) {
    let i = 0;
    while (i < n) {
        // Print the current term
        process.stdout.write(a + " ");
        a *= r;
        i++;
    }
}

// Starting number
const a = 2;

// Common ratio
const r = 3;

// Nth term to be found
const n = 5;

process.stdout.write("GP Series: ");
// Function call
printGP(a, r, n);

Output

GP Series: 2 6 18 54 162 

Time complexity: O(N) 
Auxiliary Space: O(1)