Given Mth and Nth terms of 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++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the value of thepair<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 seriesdouble 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 Codeint main(){ double m = 6, n = 10, mTerm = 12, nTerm = 20, p = 5; cout << findSum(m, n, mTerm, nTerm, p) << endl; return 0;} |
Java
// 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 Codepublic 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
# Python3 implementation of the above approachimport math as mt# Function to calculate the value of thedef 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 seriesdef 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 Codem = 6n = 10mTerm = 12nTerm = 20p = 5print(findSum(m, n, mTerm, nTerm, p))# This code is contributed by # Mohit Kumar 29 |
C#
// 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 Codepublic 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// 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?> |
Javascript
<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> |
Output:
30
Time Complexity: O(1), the code will run in a constant time.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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