Suma de P términos de un PA si se dan los términos M-ésimo y N-ésimo

Dados los términos Mth y Nth de progresión aritmética. La tarea es encontrar la suma de sus primeros p términos.
Ejemplos: 
 

Entrada: m = 6, n = 10, mth = 12, nth = 20, p = 5 
Salida: 30
Entrada: m = 10, n = 20, mth = 70, nth = 140, p = 4 
Salida: 70

Enfoque: Sea a el primer término y d la diferencia común del PA dado. Por lo tanto 
 

mth term = a + (m-1)d and
nth term = a + (n-1)d

A partir de estas dos ecuaciones, encuentre el valor de a y d. Ahora usa la fórmula de la suma de p términos de un AP. 
Suma de p términos = 
 

( p * ( 2*a + (p-1) * d ) ) / 2;

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the value of the
pair<double, double> findingValues(double m,
               double n, double mth, double nth)
{
    // Calculate value of d using formula
    double d = (abs(mth - nth)) / abs((m - 1) - (n - 1));
 
    // Calculate value of a using formula
    double a = mth - ((m - 1) * d);
 
    // Return pair
    return make_pair(a, d);
}
 
// Function to calculate value sum
// of first p numbers of the series
double findSum(int m, int n, int mth, int nth, int p)
{
 
    pair<double, double> ad;
 
    // First calculate value of a and d
    ad = findingValues(m, n, mth, nth);
 
    double a = ad.first, d = ad.second;
 
    // Calculate the sum by using formula
    double sum = (p * (2 * a + (p - 1) * d)) / 2;
 
    // Return the sum
    return sum;
}
 
// Driven Code
int main()
{
 
    double m = 6, n = 10, mTerm = 12, nTerm = 20, p = 5;
 
    cout << findSum(m, n, mTerm, nTerm, p) << endl;
 
    return 0;
}

// Java implementation of the above approach
import java.util.*;
 
class GFG
{
     
// Function to calculate the value of the
static ArrayList<Integer> findingValues(int m, int n,
                                int mth, int nth)
{
     
    // Calculate value of d using formula
    int d = (Math.abs(mth - nth)) /
        Math.abs((m - 1) - (n - 1));
 
    // Calculate value of a using formula
    int a = mth - ((m - 1) * d);
    ArrayList<Integer> res=new ArrayList<Integer>();
    res.add(a);
    res.add(d);
 
    // Return pair
    return res;
}
 
// Function to calculate value sum
// of first p numbers of the series
static int findSum(int m, int n, int mth,
                            int nth, int p)
{
    // First calculate value of a and d
    ArrayList<Integer> ad = findingValues(m, n, mth, nth);
 
    int a = ad.get(0);
    int d = ad.get(1);
 
    // Calculate the sum by using formula
    int sum = (p * (2 * a + (p - 1) * d)) / 2;
 
    // Return the sum
    return sum;
}
 
// Driver Code
public static void main (String[] args)
{
    int m = 6, n = 10, mTerm = 12, nTerm = 20, p = 5;
    System.out.println(findSum(m, n, mTerm, nTerm, p));
}
}
 
// This code is contributed by chandan_jnu

# Python3 implementation of the above approach
import math as mt
 
# Function to calculate the value of the
def findingValues(m, n, mth, nth):
 
    # Calculate value of d using formula
    d = ((abs(mth - nth)) /
          abs((m - 1) - (n - 1)))
 
    # Calculate value of a using formula
    a = mth - ((m - 1) * d)
 
    # Return pair
    return a, d
 
# Function to calculate value sum
# of first p numbers of the series
def findSum(m, n, mth, nth, p):
     
    # First calculate value of a and d
    a,d = findingValues(m, n, mth, nth)
 
    # Calculate the sum by using formula
    Sum = (p * (2 * a + (p - 1) * d)) / 2
 
    # Return the Sum
    return Sum
 
# Driver Code
m = 6
n = 10
mTerm = 12
nTerm = 20
p = 5
 
print(findSum(m, n, mTerm, nTerm, 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
static ArrayList findingValues(int m, int n,
                                int mth, int nth)
{
     
    // Calculate value of d using formula
    int d = (Math.Abs(mth - nth)) /
        Math.Abs((m - 1) - (n - 1));
 
    // Calculate value of a using formula
    int a = mth - ((m - 1) * d);
    ArrayList res=new ArrayList();
    res.Add(a);
    res.Add(d);
 
    // Return pair
    return res;
}
 
// Function to calculate value sum
// of first p numbers of the series
static int findSum(int m, int n, int mth,
                            int nth, int p)
{
    // First calculate value of a and d
    ArrayList ad = findingValues(m, n, mth, nth);
 
    int a = (int)ad[0];
    int d = (int)ad[1];
 
    // Calculate the sum by using formula
    int sum = (p * (2 * a + (p - 1) * d)) / 2;
 
    // Return the sum
    return sum;
}
 
// Driver Code
public static void Main ()
{
    int m = 6, n = 10, mTerm = 12, nTerm = 20, p = 5;
    Console.WriteLine(findSum(m, n, mTerm, nTerm, p));
}
}
 
// This code is contributed by chandan_jnu

<?php
// PHP implementation of the above approach
 
// Function to calculate the value of the
function findingValues($m, $n, $mth, $nth)
{
    // Calculate value of d using formula
    $d = (abs($mth - $nth)) /
          abs(($m - 1) - ($n - 1));
 
    // Calculate value of a using formula
    $a = $mth - (($m - 1) * $d);
 
    // Return pair
    return array($a, $d);
}
 
// Function to calculate value sum
// of first p numbers of the series
function findSum($m, $n, $mth, $nth, $p)
{
    // First calculate value of a and d
    $ad = findingValues($m, $n, $mth, $nth);
 
    $a = $ad[0];
    $d = $ad[1];
 
    // Calculate the sum by using formula
    $sum = ($p * (2 * $a + ($p - 1) * $d)) / 2;
 
    // Return the sum
    return $sum;
}
 
// Driver Code
$m = 6;
$n = 10;
$mTerm = 12;
$nTerm = 20;
$p = 5;
 
echo findSum($m, $n, $mTerm, $nTerm, $p);
 
// This code is contributed by Ryuga
?>

<script>
 
    // JavaScript implementation of the above approach
     
    // Function to calculate the value of the
    function findingValues(m, n, mth, nth)
    {
 
        // Calculate value of d using formula
        let d = parseInt(
        (Math.abs(mth - nth)) / Math.abs((m - 1) - (n - 1)), 10);
 
        // Calculate value of a using formula
        let a = mth - ((m - 1) * d);
        let res = [];
        res.push(a);
        res.push(d);
 
        // Return pair
        return res;
    }
 
    // Function to calculate value sum
    // of first p numbers of the series
    function findSum(m, n, mth, nth, p)
    {
        // First calculate value of a and d
        let ad = findingValues(m, n, mth, nth);
 
        let a = ad[0];
        let d = ad[1];
 
        // Calculate the sum by using formula
        let sum = parseInt((p * (2 * a + (p - 1) * d)) / 2, 10);
 
        // Return the sum
        return sum;
    }
     
    let m = 6, n = 10, mTerm = 12, nTerm = 20, p = 5;
    document.write(findSum(m, n, mTerm, nTerm, p));
     
</script>

30

Complejidad de tiempo: O(1), el código se ejecutará en un tiempo constante.
Espacio auxiliar: O(1), no se requiere espacio adicional, por lo que es una constante.