Sum of P terms of an AP if Mth and Nth terms are given

Given Mth and Nth term of an arithmetic progression. The task is to find the sum of its first p terms.

Examples:

Input: m = 6, n = 10, mth = 12, nth = 20, p = 5
Output:30

Input:m = 10, n = 20, mth = 70, nth = 140, p = 4
Output:70

Approach: Let a is the first term and d is the common difference of the given AP. Therefore

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

From these two equations, find the value of a and d. Now use the formula of sum of p terms of an AP.
Sum of p terms =

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

Below is the implementation of the above approach:

C++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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Output:

30


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