Find Pth term of a GP if Mth and Nth terms are given

Given Mth and Nth term of a Geometric progression. Find its Pth term.
Examples:

Input: m = 10, n = 5, mth = 2560, nth = 80, p = 30
Output: pth = 81920
Input: m = 8, n = 2, mth = 1250, nth = 960, p = 15
Output: 24964.4

Approach:
Let a is the first term and r is the common ratio of the given Geometric Progression. Therefore

mth term = a * pow ( r, (m-1) ) ....... (i) and
nth term = a * pow ( r, (n-1) ) ....... (ii)

For convenience, it is assumed that m > n
From these 2 equations,
Since we have given values m, n, mth term, and nth term, therefore

r = pow(A/B, 1.0/(m-n))

and
Now put the value of r in any of above two-equation and calculate the value of a.

a = mth term / pow ( r, (m-1) ) or
a = nth term / pow ( r, (n-1) )

After finding the value of a and r, use the formula of Pth terms of a GP.

pth term of GP = a * pow ( r, (p-1.0) );

Below is the implementation of the above approach:

C++

#include <cmath>
#include <iostream>
#include <vector>

using namespace std;
 
// function to calculate the value
// of the a and r of geometric series

pair<double, double> values_of_r_and_a(double m,

                                       double n,

                                       double mth,

                                       double nth)
{

    double a, r;
 
    if (m < n) {

        swap(m, n);

        swap(mth, nth);

    }
 
    // calculate value of r using formula

    r = pow(mth / nth, 1.0 / (m - n));
 
    // calculate value of a using value of r

    a = mth / pow(r, (m - 1));
 
    // push both values in the vector and return it

    return make_pair(a, r);
}
 
// function to calculate the value
// of pth term of the series

double FindSum(int m, int n, double mth,

               double nth, int p)
{

    pair<double, double> ar;
 
    // first calculate value of a and r

    ar = values_of_r_and_a(m, n, mth, nth);
 
    double a = ar.first;

    double r = ar.second;
 
    // calculate pth term by using formula

    double pth = a * pow(r, (p - 1.0));
 
    // return the value of pth term

    return pth;
}
 
// Driven program to test

int main()
{

    int m = 10, n = 5, p = 15;

    double mth = 2560, nth = 80;

    cout << FindSum(m, n, mth, nth, p)

         << endl;
 
    return 0;
}

Java

// Java implementation of the above approach

import java.util.ArrayList;
 
class GFG
{
 
// function to calculate the value 
// of the a and r of geometric series 

static ArrayList values_of_r_and_a(double m, double n,

                                double mth, double nth) 
{ 

    if (m < n)

    { 

        double t = m;

        n = m;

        m = t;

        t = mth;

        mth = nth;

        nth = t;

    } 
 
    // calculate value of r using formula 

    double r = Math.pow(mth / nth, 1.0 / (m - n)); 
 
    // calculate value of a using value of r 

    double a = mth / Math.pow(r, (m - 1)); 
 
    // push both values in the vector 

    // and return it

    ArrayList arr = new ArrayList();

    arr.add(a);

    arr.add(r);

    return arr; 
} 
 
// function to calculate the value 
// of pth term of the series 

static double FindSum(double m, double n, 

                    double mth, double nth,

                    double p) 
{ 
 
    // first calculate value of a and r 

    ArrayList ar = values_of_r_and_a(m, n, mth, nth); 
 
    double a = (double)ar.get(0); 

    double r = (double)ar.get(1); 
 
    // calculate pth term by using formula 

    double pth = a * Math.pow(r, (p - 1.0)); 
 
    // return the value of pth term 

    return pth; 
} 
 
// Driver Code

public static void main(String[] args)
{

    double m = 10;

    double n = 5;

    double p = 15; 

    double mth = 2560;

    double nth = 80;
 
    System.out.println((int)FindSum(m, n, mth, nth, p));
}
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 program for above approach
 
# function to calculate the value
# of the a and r of geometric series

def values_of_r_and_a(m, n, mth, nth):
 
    a, r = 0.0, 0.0
 
    if (m < n):

        m, n = n, m

        mth, nth = mth, nth
 
    # calculate value of r using formula

    r = pow(mth // nth, 1.0 /(m - n))
 
    # calculate value of a using value of r

    a = mth // pow(r, (m - 1))
 
    # push both values in the vector

    # and return it

    return a, r
 
# function to calculate the value
# of pth term of the series

def FindSum(m, n, mth, nth, p):
 
    # first calculate value of a and r

    a,r = values_of_r_and_a(m, n, mth, nth)
 
    # calculate pth term by using formula

    pth = a * pow(r, (p - 1.0))
 
    # return the value of pth term

    return pth
 
# Driven Code

m, n, p = 10, 5, 15

mth, nth = 2560.0, 80.0

print(FindSum(m, n, mth, nth, p))

# This code is contributed by 
# Mohit kumar 29

// C# implementation of the above approach

using System;

using System.Collections;
 
class GFG
{
 
// function to calculate the value 
// of the a and r of geometric series 

static ArrayList values_of_r_and_a(double m, double n,

                                double mth, double nth) 
{ 

    if (m < n)

    { 

        double t = m;

        n = m;

        m = t;

        t = mth;

        mth = nth;

        nth = t;

    } 
 
    // calculate value of r using formula 

    double r = Math.Pow(mth / nth, 1.0 / (m - n)); 
 
    // calculate value of a using value of r 

    double a = mth / Math.Pow(r, (m - 1)); 
 
    // push both values in the vector 

    // and return it

    ArrayList arr = new ArrayList();

    arr.Add(a);

    arr.Add(r);

    return arr; 
} 
 
// function to calculate the value 
// of pth term of the series 

static double FindSum(double m, double n, 

                    double mth, double nth,

                    double p) 
{ 
 
    // first calculate value of a and r 

    ArrayList ar = values_of_r_and_a(m, n, mth, nth); 
 
    double a = (double)ar[0]; 

    double r = (double)ar[1]; 
 
    // calculate pth term by using formula 

    double pth = a * Math.Pow(r, (p - 1.0)); 
 
    // return the value of pth term 

    return pth; 
} 
 
// Driver Code

static void Main()
{

    double m = 10;

    double n = 5;

    double p = 15; 

    double mth = 2560;

    double nth = 80;
 
    Console.WriteLine(FindSum(m, n, mth, nth, p));
}
}
 
// This code is contributed by mits

PHP

<?php
// Php implementation of the above approach

function swap($a1, $a2)
{

    $temp = $a1;

    $a1 = $a2;

    $a2 = $temp;
}
 
// function to calculate the value 
// of the a and r of geometric series 

function values_of_r_and_a($m, $n, $mth, $nth) 
{ 

    if ($m < $n)

    { 

        swap($m, $n); 

        swap($mth, $nth); 

    } 
 
    // calculate value of r using formula 

    $r = pow($mth / $nth, 1.0 / ($m - $n)); 
 
    // calculate value of a using value of r 

    $a = $mth / pow($r, ($m - 1)); 
 
    // push both values in the vector 

    // and return it 

    return array($a, $r); 
} 
 
// function to calculate the value 
// of pth term of the series 

function FindSum($m, $n, $mth, $nth, $p) 
{ 
 
    // first calculate value of a and r 

    $ar = values_of_r_and_a($m, $n, $mth, $nth); 
 
    $a = $ar[0]; 

    $r = $ar[1]; 
 
    // calculate pth term by using formula 

    $pth = $a * pow($r, ($p - 1.0)); 
 
    // return the value of pth term 

    return $pth; 
} 
 
// Driver Code

$m = 10;

$n = 5;

$p = 15; 
 
$mth = 2560;

$nth = 80;
 
echo FindSum($m, $n, $mth, $nth, $p);
 
// This code is contributed by Ryuga
?>

Javascript

<script>

    // Javascript implementation of the above approach

    // function to calculate the value 

    // of the a and r of geometric series 

    function values_of_r_and_a(m, n, mth, nth) 

    { 

        if (m < n)

        { 

            let t = m;

            n = m;

            m = t;

            t = mth;

            mth = nth;

            nth = t;

        } 
 
        // calculate value of r using formula 

        let r = Math.pow(mth / nth, 1.0 / (m - n)); 
 
        // calculate value of a using value of r 

        let a = mth / Math.pow(r, (m - 1)); 
 
        // push both values in the vector 

        // and return it

        let arr = [];

        arr.push(a);

        arr.push(r);

        return arr; 

    } 
 
    // function to calculate the value 

    // of pth term of the series 

    function FindSum(m, n, mth, nth, p) 

    { 
 
        // first calculate value of a and r 

        let ar = values_of_r_and_a(m, n, mth, nth); 
 
        let a = ar[0]; 

        let r = ar[1]; 
 
        // calculate pth term by using formula 

        let pth = a * Math.pow(r, (p - 1.0)); 
 
        // return the value of pth term 

        return pth; 

    } 

    let m = 10;

    let n = 5;

    let p = 15; 

    let mth = 2560;

    let nth = 80;

    document.write(FindSum(m, n, mth, nth, p));

</script>

Output:

Time Complexity: O(log₂m + log₂p), where m and p represents the value of the given integers.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Article Tags :

C++ Programs

DSA

Mathematical

School Programming

Geometric Progression